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CHAPTER 8
8
INFINITE SERIES
Thus far in our study of calculus, we have developed three major mathematical concepts:
the limit, the derivative and the definite integral. As you recall, the derivative and the
definite integral are each defined in terms of a limit. In this chapter, we explore a new
concept, called a series (or infinite series), defined in terms of (what else?) a limit. Unlike
the derivative and the definite integral, which were motivated by trying to calculate the
physical quantities of velocity and area, respectively, we have no immediate physical
motivation for the development of series. However,
infinite series are essential to countless important
applications.
You have probably had the experience of turning
on a favorite music CD and adjusting the bass and treble controls until you got the sound just right. While
your primary concern here is with getting a solid bass feeling or a brighter sound, you may
be interested to learn that there is a great deal of physics and engineering that goes into the
design of musical equipment. What we hear as a stinging guitar or soulful saxophone is
actually a precise combination of sounds of many different frequencies. Modern stereos
and music synthesizers are essentially special purpose computers designed to reproduce
complex musical waveforms with maximum fidelity. Your stereo adjustments alter the
combination of high-frequency and low-frequency vibrations with which your speakers
produce sound.
The mathematical problem we explore in this chapter is closely connected to the
musical problem described above. We want to find a combination of basic elements (in this
case, certain simple functions) that produce a desired result. Our need is a practical one: we
need to compute with an ever-expanding list of functions. Unfortunately, our ability to do
so is extremely limited for transcendental functions (e.g., sin x, cos x, ln x, e x , etc.). In
fact, let’s be completely honest: we just don’t know how to compute values of these
functions very well. For instance, how would you compute sin(1.234567)? Yes, we’re
aware that any scientific calculator can give you an answer, but that’s not really the question. For a polynomial, say
f (x) = 6x 4 − 15x 3 + 3x 2 +
12
x − 142,
7
you can compute the value of f (x), for any x, simply by performing the indicated arithmetic. But, there is no arithmetic for computing sin(1.234567). So, how is it done, then?
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Chapter 8 Infinite Series
Over the course of this chapter, we will develop the mathematical machinery you need to
calculate highly accurate approximations to any transcendental function. This is no small
matter, as you will see, but the benefits of having infinite series available are substantial.
Series are used widely in applications, such as the solution of differential equations. Constructing series (specifically, power series) will yield an explosion of new transcendental
functions. In fact, many functions of significance in physics and engineering (e.g., Bessel
functions) are defined as series.
Music synthesis and the evaluation of transcendental functions are only two of countless applications of infinite series. In our increasingly digital world, you can find these
ideas everywhere: digital television, digital photography and so on, as well as more traditional scientific research areas such as holography, tomography and spectroscopy all rely
heavily on the concept of infinite series. Although it takes us six sections in this chapter to
develop the mathematical tools necessary to understand infinite series, stick with us. The
applications in the last three sections (and in future courses) are well worth your effort!
8.1
Definition of sequence
SEQUENCES OF REAL NUMBERS
The mathematical notion of sequence is not much different from the common English usage
of the word. For instance, if you were asked to describe the sequence of events that led up
to a traffic accident, you’d not only need to list the events, but you’d need to do so in a specific order (hopefully, the order in which they actually occurred). In mathematics, we use
the term sequence to mean an infinite collection of real numbers, written in a specific order.
We have already seen sequences several times now (although we have not formally introduced the notion). For instance, you have found approximate solutions to nonlinear equations like tan x − x = 0, by first making an initial guess, x0 and then using Newton’s method
to compute a sequence of successively improved approximations, x1 , x2 , . . . , xn , . . . .
By sequence we mean any function whose domain is the set of integers starting with
some integer n 0 (often 0 or 1). For instance, the function a(n) = n1 , for n = 1, 2, 3, . . . ,
defines the sequence
1 1 1 1
, , , ,....
1 2 3 4
Here, 11 is called the first term, 12 is the second term and so on. We call a(n) = n1 the
general term, since it gives a (general) formula for computing all the terms of the sequence. Further, we usually use subscript notation instead of function notation and write an
instead of a(n).
Example 1.1
The Terms of a Sequence
Write out the terms of the sequence whose general term is given by an =
n = 1, 2, 3, . . . .
Solution
We have the sequence
a1 =
■
1+1
2
2+1
3
4
5
= , a2 =
= , a3 = , a4 = , . . . .
1
1
2
2
3
4
n+1
, for
n
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Section 8.1
an
Sequences of Real Numbers
623
We often use set notation to denote a sequence. For instance, the sequence with gen1
eral term an = 2 , for n = 1, 2, 3, . . . , is denoted by
n
∞
1
∞
{an }n=1 =
,
n 2 n=1
.12
.08
or equivalently, by listing the terms of the sequence:
1 1 1
1
, , ,..., 2,... .
1 22 32
n
.04
n
5
10
15
Figure 8.1
1
an = 2 .
n
20
To graph this sequence, we plot a number of discrete points, since a sequence is a function
defined only on the integers
∞ (see Figure 8.1). You have likely already noticed something
1
. As n gets larger and larger, the terms of the sequence,
about the sequence
n 2 n=1
1
an = 2 get closer and closer to zero. In this case, we say that the sequence converges to
n
0 and write
lim an = lim
n→∞
n→∞
1
= 0.
n2
lim an = L if we
In general, we say that the sequence {an }∞
n=1 converges to L i.e., n→∞
can make an as close to L as desired, simply by making n sufficiently large. You may notice
that this language parallels that used in the definition of the limit
lim f (x) = L
x→∞
for a function of a real variable x (given in section 1.5). The only difference is that n can
take on only integer values, while x can take on any real value (integer, rational or
irrational).
When we say that we can make an as close to L as desired (i.e., arbitrarily close), just
how close must we be able to make an to L? Well, if you pick any (small) real number, ε > 0,
you must be able to make an within a distance ε of L , simply by making n sufficiently large.
Think about this carefully; an is within distance ε of L if
L − ε < an < L + ε.
This is equivalent to
−ε < an − L < ε
or |an − L| < ε.
We summarize this in the following definition.
Definition 1.1
The sequence {an }∞
n=n 0 converges to L if and only if given any number ε > 0, there is
an integer N for which
|an − L| < ε, for every n > N.
If there is no such number L , then we say that the sequence diverges.
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Chapter 8 Infinite Series
We illustrate Definition 1.1 in Figure 8.2. Notice that the definition says that the
sequence {an }∞
n=1 converges to L if given any number ε > 0, we can find an integer N , so
that the terms of the sequence stay between L − ε and L + ε for all values of n > N.
y
L´
L
L´
n
12345
N
Figure 8.2
Convergence of a sequence.
In the following example, we show how to use Definition 1.1 to prove that a sequence
converges.
Example 1.2
Proving That a Sequence Converges
∞
1
Use Definition 1.1 to show that the sequence
converges to 0.
n 2 n=1
1
Solution
In this case, we must show that we can make 2 as close to 0 as desired,
n
just by making n sufficiently large. So, given any ε > 0, we must find N sufficiently
large so that for every n > N,
1
− 0 < ε.
n2
Notice that this is equivalent to
1
< ε.
n2
(1.1)
Observe that since n 2 and ε are positive, we can divide both sides of (1.1) by ε and
multiply by n 2 , to obtain
1
< n2.
ε
Taking square roots gives us
1
< n.
ε
So (since each
of the preceeding steps is reversible), if we choose N to be an integer
1
1
, then n > N implies that 2 < ε, as desired.
with N ≥
ε
n
■
Most of the usual rules for computing limits of functions of a real variable also apply
to computing the limit of a sequence, as we see in the following result.
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Section 8.1
Sequences of Real Numbers
625
Theorem 1.1
∞
Suppose that {an }∞
n=n 0 and {bn }n=n 0 both converge. Then
an
(i)
6
lim (an + bn ) = lim an + lim bn ,
n→∞
n→∞
n→∞
lim (an − bn ) = lim an − lim bn ,
n→∞
n→∞
n→∞
lim bn and
(iii) lim (an bn ) = lim an
(ii)
5
4
n→∞
3
n→∞
n→∞
lim an
2
(iv)
1
lim
n→∞
an
n→∞
=
(assuming lim bn = 0).
n→∞
bn
lim bn
n→∞
n
5
10
15
20
The proof of Theorem 1.1 is virtually identical to the proof of the corresponding
theorem about limits of a function of a real variable (see Theorem 2.3 in section 1.2 and
Appendix A) and is omitted.
Figure 8.3
5n + 7
an =
.
3n − 5
Remark 1.1
To find the limit of a sequence, you should work largely the same as when computing the limit of a
function of a real variable, but keep in mind that sequences are defined only for integer values of the
variable.
NOTES
If you (incorrectly) apply
l’Hôpital’s Rule in example 1.3,
you get the right answer. (Go ahead
and try it; nobody’s looking.)
Unfortunately, you will not always
be so lucky. It’s a lot like trying to
cross a busy highway: while there
are times when you can
successfully cross with your eyes
closed, it’s not generally
recommended. Theorem 1.2
describes how you can safely use
l’Hôpital’s Rule.
Example 1.3
Finding the Limit of a Sequence
5n + 7
.
n→∞ 3n − 5
Evaluate lim
12
∞.
Solution
Of course, this has the indeterminate form
From the graph in
∞
Figure 8.3, it looks like the sequence tends to some limit around 2. Note that we cannot
apply l’Hôpital’s Rule here, since the functions in the numerator and the denominator
are not continuous. (They are only defined for integer values of n, even though you
could define these expressions for any real values of n.) You can of course, use the simpler method of dividing numerator and denominator by the highest power of n in the
denominator. We have
(5n + 7) n1
5 + n7
5n + 7
5
1 = lim
lim
= lim
= .
n→∞ 3n − 5
n→∞ (3n − 5)
n→∞ 3 − 5
3
n
n
10
■
an
8
In the following example, we see a sequence that diverges by virtue of its terms
tending to +∞.
6
4
Example 1.4
2
n
5
10
15
Figure 8.4
n2 + 1
an =
.
2n − 3
20
A Divergent Sequence
n2 + 1
.
n→∞ 2n − 3
Evaluate lim
∞,
Solution
Again, this has the indeterminate form
but from the graph in
∞
Figure 8.4, the sequence appears to be increasing without bound. Dividing top and
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Chapter 8 Infinite Series
an
bottom by n (the highest power of n in the denominator) we have
(n 2 + 1) n1
n + n1
n2 + 1
1 = lim
lim
=∞
= lim
n→∞ 2n − 3
n→∞ (2n − 3)
n→∞ 2 − 3
n
n
2
∞
n +1
diverges.
and so, the sequence
2n − 3 n=1
1
n
5
10
15
1
■
Figure 8.5
an = (−1)n .
In the following example, we see that a sequence doesn’t need to tend to ±∞ in order
to diverge.
an
Example 1.5
2
A Divergent Sequence Whose Terms Do Not Tend to ∞
Determine the convergence or divergence of the sequence {(−1)n }∞
n=1 .
1
Solution
If we write out the terms of the sequence, we have
n
5
{−1, 1, −1, 1, −1, 1, . . .}.
10
Figure 8.6
an = f (n), where f (x) → 2, as
x → ∞.
That is, the terms of the sequence alternate back and forth between −1 and 1 and so, the
sequence diverges. To see this graphically, we plot the first few terms of the sequence in
Figure 8.5. Notice that the points do not approach any limit (a horizontal line).
an
■
1.2
You can use an advanced tool like l’Hôpital’s Rule to find the limit of a sequence, but
you must be careful. The following theorem says that if f (x) → L as x → ∞ through
all real values, then f (n) must approach L , too, as n → ∞ through integer values. (See
Figure 8.6 for a graphical representation of this.)
1.0
0.8
0.6
0.4
0.2
n
5
10
15
20
Theorem 1.2
Suppose that lim f (x) = L . Then, lim f (n) = L , also.
x→∞
Figure 8.7a
an = cos(2πn).
n→∞
y
Remark 1.2
1
The converse of Theorem 1.2 is false. That is, if lim f (n) = L , it need not be true that lim f (x) = L .
n→∞
0.5
This is clear from the following observation. Note that
lim cos(2πn) = 1,
x
n→∞
10
since cos(2πn) = 1 for every integer n (see Figure 8.7a).
However,
0.5
1
lim cos(2π x) does not exist,
x→∞
Figure 8.7b
y = cos(2π x).
since as x → ∞, cos(2π x) oscillates between −1 and 1 (see Figure 8.7b).
x →∞
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Section 8.1
Example 1.6
Sequences of Real Numbers
627
Applying L’Hôpital’s Rule to a Related Function
n+1
.
n→∞ en
an
Evaluate lim
0.8
0.6
0.4
0.2
n
5
10
15
20
Figure 8.8
n+1
an =
.
en
∞,
Solution
This has the indeterminate form
but from the graph in Figure 8.8, it
∞
appears that the sequence converges to 0. However, there is no obvious way to resolve
this, except by l’Hôpital’s Rule (which does not apply to limits of sequences). So, we instead consider the limit of the corresponding function of a real variable to which we may
apply l’Hôpital’s Rule. (Be sure you check the hypotheses.) We have
d
(x + 1)
x +1
1
dx
= lim x = 0.
lim
=
lim
d x
x→∞ e x
x→∞
x→∞ e
(e )
dx
From Theorem 1.2, we now have
n+1
= 0, also.
lim
n→∞ en
■
Although we now have a few tools for computing the limit of a sequence, most interesting sequences resist our attempts to find their limit. In many cases (including infinite
series, which we study in the remainder of this chapter), we don’t even have an explicit
formula for the general term. In such circumstances, we must test the sequence for convergence in some indirect way. The first indirect tool we present corresponds to the result (of
the same name) for limits of functions of a real variable presented in section 1.2.
Theorem 1.3 (Squeeze Theorem)
∞
Suppose {an }∞
n=n 0 and {bn }n=n 0 are convergent sequences, both converging to the limit, L.
If there is an integer n 1 ≥ n 0 such that for all n ≥ n 1 , an ≤ cn ≤ bn , then {cn }∞
n=n 0 converges to L, too.
In the following example, we demonstrate how to apply the Squeeze Theorem to a
sequence. Observe that the trick here is to find two sequences, one on either side of the
given sequence (i.e., one larger and one smaller) that have the same limit.
an
Example 1.7
Applying the Squeeze Theorem to a Sequence
sin n ∞
.
Determine the convergence or divergence of
n 2 n=1
0.25
0.20
0.15
0.10
0.05
n
0.05
5
10
15
Figure 8.9
sin n
an = 2 .
n
20
Solution
From the graph in Figure 8.9, the sequence appears to converge to 0,
despite the oscillation. Further, note that you cannot compute this limit using the rules
we have established so far. (Try it!) However, recall that
−1 ≤ sin n ≤ 1, for all n.
Dividing through by n 2 gives us
sin n
1
−1
≤ 2 ≤ 2 , for all n ≥ 1.
2
n
n
n
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Chapter 8 Infinite Series
Finally, observe that
−1
1
= 0 = lim 2 .
2
n→∞
n
n
lim
n→∞
From the Squeeze Theorem, we now have that
lim
n→∞
sin n
= 0,
n2
also.
■
The following useful result follows immediately from Theorem 1.3.
Corollary 1.1
If lim |an | = 0, then lim an = 0, also.
n→∞
n→∞
Proof
Notice that for all n,
−|an | ≤ an ≤ |an |
and
lim |an | = 0 and
n→∞
lim (−|an |) = − lim |an | = 0.
n→∞
n→∞
So, from the Squeeze Theorem, lim an = 0, too.
n→∞
■
Corollary 1.1 is particularly useful for sequences with both positive and negative
terms, as in the following example.
Example 1.8
A Sequence with Terms of Alternating Signs
(−1)n ∞
.
Determine the convergence or divergence of
n
n=1
an
0.5
n
5
10
15
0.5
20
Solution
From the graph of the sequence in Figure 8.10, it seems that the sequence
oscillates but still may be converging to 0. Since (−1)n oscillates back and forth
(−1)n
between −1 and 1, we cannot compute lim
directly. However, notice that
n→∞
n
(−1)n 1
n = n
and
1
lim
n→∞
Figure 8.10
(−1)n
an =
.
n
1
= 0.
n
(−1)n
= 0, too.
n→∞
n
From Corollary 1.1, we get that lim
■
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Section 8.1
629
Sequences of Real Numbers
We remind you of the following definition, which we use throughout the chapter.
Definition 1.2
For any integer n ≥ 1, the factorial, n! is defined as the product of the first n positive
integers,
n! = 1 · 2 · 3 · · · · · n.
We define 0! = 1.
The following example shows a sequence whose limit would be extremely difficult to
find without the Squeeze Theorem.
Example 1.9
An Indirect Proof of Convergence
∞
n!
.
Investigate the convergence of
n n n=1
n!
Solution
First, notice that we have no means of computing lim n directly. (Try
n→∞ n
this!) From the graph of the sequence in Figure 8.11, it appears that the sequence is
converging to 0. Notice that the general term of the sequence satisfies
n!
1 · 2 · 3 · ··· · n
0< n =
n
n
· n · n· · · · · n
an
1.0
0.8
0.6
n factors
0.4
1 2 · 3 · ··· · n
1
1
=
≤
(1) = .
n n · n ·· · · · n
n
n
0.2
n
5
10
15
Figure 8.11
n!
an = n .
n
(1.2)
n − 1 factors
20
From the Squeeze Theorem and (1.2), we have that since
lim
n→∞
1
= 0 and
n
lim 0 = 0,
n→∞
then
lim
n→∞
n!
= 0, also.
nn
■
Just as we did with functions of a real variable, we need to distinguish between
sequences that are increasing and decreasing. The definitions are quite straightforward.
Definition 1.3
(i) The sequence {an }∞
n=1 is increasing if
a1 ≤ a2 ≤ · · · ≤ an ≤ an+1 ≤ · · · .
(ii)
The sequence {an }∞
n=1 is decreasing if
a1 ≥ a2 ≥ · · · ≥ an ≥ an+1 ≥ · · · .
If a sequence is either increasing or decreasing, it is called monotonic.
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There are any number of ways to show that a sequence is monotonic. Regardless of
which method you use, you will need to show that either an ≤ an+1 for all n (increasing) or
an+1 ≤ an for all n (decreasing). One very useful method is to look at the ratio of the two
successive terms an and an+1 . We illustrate this in the following example.
Example 1.10
An Increasing Sequence
∞
n
Investigate whether the sequence
is increasing, decreasing or neither.
n + 1 n=1
Solution
From the graph in Figure 8.12, it appears that the sequence is increasing.
However, you should not be deceived by looking at the first few terms of a sequence.
n ,
Instead, we look at the ratio of two successive terms. So, if we define an =
we
n
+
1
n+1
and
have an+1 =
n+2
n+1
an+1
n+1 n+1
n+2
=
=
n
an
n+2 n
n+1
an
1.0
0.8
0.6
0.4
0.2
n
5
10
15
20
=
Figure 8.12
n
an =
.
n+1
n 2 + 2n + 1
1
=1+ 2
> 1.
2
n + 2n
n + 2n
(1.3)
Since an > 0, notice that we can multiply both sides of (1.3) by an , to obtain
an+1 > an ,
for all n and so, the sequence is increasing. As an alternative, notice that you can always
x
consider the function f (x) =
(of the real variable x) corresponding to the
x +1
sequence. Observe that
f (x) =
(x + 1) − x
1
=
> 0,
2
(x + 1)
(x + 1)2
which says that the function f (x) is increasing. From this, it follows that the corren
sponding sequence an =
is also increasing.
n+1
■
A Sequence That Is Increasing for n ≥ 2
∞
n!
Investigate whether the sequence n
is increasing, decreasing or neither.
e n=1
Example 1.11
an
200
150
Solution
From the graph of the sequence in Figure 8.13, it appears that the sequence
(n + 1)!
n!
,
is increasing (and rather rapidly, at that). Here, for an = n , we have an+1 =
e
en+1
so that
100
50
n
5
Figure 8.13
n!
an = n .
e
10
an+1
an
(n + 1)!
(n + 1)! en
en+1
= =
n!
en+1 n!
n
e
n+1
(n + 1)n! en
=
> 1, for n ≥ 2.
=
n
e(e )n!
e
Since (n + 1)! = (n + 1) · n!
and en+1 = e · en.
(1.4)
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Section 8.1
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631
Multiplying both sides of (1.4) by an > 0, we get
an+1 > an , for n ≥ 2.
Notice that in this case, although the sequence is not increasing for all n, it is increasing
for n ≥ 2. Keep in mind that it doesn’t really matter what the first few terms do, anyway.
We are only concerned with the behavior of a sequence as n → ∞.
■
We need to define one additional property of sequences.
Definition 1.4
We say that the sequence {an }∞
n=n 0 is bounded if there is a number M > 0 (called a
bound) for which |an | ≤ M, for all n.
There is often some slight confusion here. A bound is not the same as a limit, although
the two may coincide. The limit of a convergent sequence is the value that the terms of the
sequence are approaching as n → ∞. On the other hand, a bound is any number that is
greater than or equal to the absolute value of every term. This says that a given sequence may
have any number of bounds (e.g., if |an | ≤ 10 for all n, then |an | ≤ 20, for all n, too). However, a sequence may have only one limit (or no limit, in the case of a divergent sequence).
Example 1.12
A Bounded Sequence
∞
3 − 4n 2
Show that the sequence
is bounded.
n 2 + 1 n=1
an
1.0
0.8
Solution
0.6
0.4
0.2
n
5
10
15
20
Figure 8.14a
A bounded and increasing sequence.
We use the fact that 4n 2 − 3 > 0, for all n ≥ 1, to get
3 − 4n 2 4n 2 − 3
4n 2
4n 2
=
|an | = 2
<
<
= 4.
n +1 n2 + 1
n2 + 1
n2
So, this sequence is bounded by 4. (We might also say in this case that the sequence is
bounded between −4 and 4.) Further, note that we could also use any number greater
than 4 as a bound.
■
We should emphasize that the reason we are considering whether a sequence is
monotonic or bounded is that very often we cannot compute the limit of a given sequence
directly and must rely on indirect means to determine whether or not the sequence is
convergent. The following theorem provides a powerful tool for the investigation of
sequences.
an
6
4
Theorem 1.4
Every bounded, monotonic sequence converges.
2
n
5
10
15
20
Figure 8.14b
A bounded and decreasing
sequence.
A typical bounded and increasing sequence is illustrated in Figure 8.14a, while a
bounded and decreasing sequence is illustrated in Figure 8.14b. In both figures, notice that
a bounded and monotonic sequence has nowhere to go and consequently, must converge.
The proof of Theorem 1.4 is rather involved and we leave it to the end of the section.
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Chapter 8 Infinite Series
an
In the very common case where we do not know how to compute the limit of a
sequence, this theorem says that if we can show that a sequence is bounded and monotonic, then it must also be convergent, although we may have little idea of what its limit
might be. Once we establish that a sequence converges, we can approximate its limit by
computing a sufficient number of terms, as in the following example.
2.0
1.5
1.0
Example 1.13
0.5
n
5
10
15
20
Figure 8.15
2n
an = .
n!
an =
n
2n
n!
2
2
4
0.666667
6
0.088889
8
0.006349
10
0.000282
12
0.0000086
14
1.88 × 10−7
16
3.13 × 10−9
18
4.09 × 10−11
20
4.31 × 10−13
Remark 5.3
Do not underestimate the importance
of Theorem 1.4. This indirect way of
testing a sequence for convergence
takes on additional significance
when we study infinite series (a special type of sequence that is the topic
of the remainder of this chapter).
An Indirect Proof of Convergence
n ∞
2
.
Investigate the convergence of the sequence
n! n=1
2n
. This has the
First, note that we do not know how to compute lim
n→∞ n!
∞,
but we cannot apply l’Hôpital’s Rule to it. (Why not?) The
indeterminate form
∞
graph in Figure 8.15 suggests that the sequence converges to some number close to 0.
To confirm this suspicion, we first show that the sequence is monotonic. We have
n+1 2
an+1
2n+1 n!
(n + 1)!
= n =
2
an
(n + 1)! 2n
n!
Solution
=
2(2n ) n!
2
=
≤ 1, for n ≥ 1.
n
(n + 1) n! 2
n+1
Since 2n+1 = 2 · 2n and
(n + 1)! = (n + 1) · n!
(1.5)
Multiplying both sides of (1.5) by an gives us an+1 ≤ an , for all n and so, the sequence
is decreasing. Next, since we have already shown that the sequence is decreasing,
observe that
0<
2n
21
≤
= 2,
n!
1!
for n ≥ 1 (i.e., the sequence is bounded by 2). Since the sequence is both bounded and
monotonic, it must be convergent, by Theorem 1.4. To get an approximation of the limit
of the sequence, we display a number of terms of the sequence in the accompanying table.
From the table, it appears that the sequence is converging to approximately 0. We
can make a slightly stronger statement, though. Since we have established that the
sequence is decreasing and convergent, we have from our computations that
0 ≤ an ≤ a20 ≈ 4.31 × 10−13 , for n ≥ 20.
Further, the limit L must also satisfy the inequality
0 ≤ L ≤ 4.31 × 10−13 .
So, even if we can’t conclude that the sequence converges to 0, we can conclude that it
converges to some number extremely close to zero. For most purposes, such an estimate
is entirely adequate. Of course, if you need greater precision, you can always compute
a few more terms.
■
Proof of Theorem 1.4
Before we can prove Theorem 1.4, we need to state one of the properties of the real number system.
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Section 8.1
Sequences of Real Numbers
633
The Completeness Axiom
If a nonempty set S of real numbers has a lower bound, then it has a greatest lower
bound. Equivalently, if it has an upper bound, it has a least upper bound.
This axiom says that if a nonempty set S has an upper bound, that is, a number M such
that
x ≤ M, for all x ∈ S,
then there is an upper bound L, for which
L ≤ M for all upper bounds, M,
with a corresponding statement holding for lower bounds.
The Completeness Axiom enables us to prove Theorem 1.4.
Proof
(Increasing sequence) Suppose that {an }∞
n=n 0 is increasing and bounded. This is illustrated in Figure 8.16, where you can see an increasing sequence bounded by 1. We have
an
a1 ≤ a2 ≤ a3 ≤ · · · ≤ an ≤ an+1 ≤ · · ·
1.0
0.8
and for some number M > 0, |an | ≤ M for all n. This is the same as saying that
0.6
−M ≤ an ≤ M, for all n.
0.4
0.2
n
5
10
15
20
Figure 8.16
Bounded and increasing.
Now, let S be the set containing all of the terms of the sequence, S = {a1 , a2 , . . . ,
an , . . .}. Notice that M is an upper bound for the set S. From the Completeness Axiom,
S must have a least upper bound, L. That is, L is the smallest number for which
an ≤ L , for all n.
(1.6)
Notice that for any number ε > 0, L − ε < L and so, L − ε is not an upper bound,
since L is the least upper bound. Since L − ε is not an upper bound for S, there is some
element, a N of S for which
L − ε < aN .
Since {an } is increasing, we have that for n ≥ N, a N ≤ an . Finally, from (1.6) and the
fact that L is an upper bound for S and since ε > 0, we have
L − ε < a N ≤ an ≤ L < L + ε,
or more simply
L − ε < an < L + ε,
for n ≥ N. This is equivalent to
|an − L| < ε, for n ≥ N.
This says that {an } converges to L. The proof for the case of a decreasing sequence is
similar and is left as an exercise.
■
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Chapter 8 Infinite Series
EXERCISES 8.1
Compare and contrast lim sin π x and lim sin πn.
1.
x →∞
n→∞
21. an = (−1)n
n+2
3n − 1
22. an = (−1)n
n+4
n+1
23. an = (−1)n
n+2
n2 + 4
24. an = (−1)n
4
n+1
Indicate the domains of the two functions and how this
affects the limits.
2.
Explain why Theorem 1.2 should be true, taking into
account the respective domains and their effect on the
limits.
3.
In words, explain why a decreasing bounded sequence
must converge.
4.
A sequence is said to diverge if it does not converge. The
word “diverge’’ is well chosen for sequences that
diverge to ∞, but is less descriptive of sequences such as
{1, 2, 1, 2, 1, 2, . . .} and {1, 2, 3, 1, 2, 3, . . .}. Briefly describe
the limiting behavior of these sequences and discuss other
possible limiting behaviors of divergent sequences.
In exercises 5–8, write out the terms a1 , a2 , . . . , a6 of the given
sequence.
2n − 1
5. an =
n2
7. an =
4
n!
11.
an =
n
n+1
2
13. an = √
n
12. an =
28. an =
cos n
en
en + 2
e2n − 1
30. an =
e2n
en + 1
3n
+1
32. an =
n2n
3n
34. an =
n!
2n
29. an =
31. an =
33. an =
en
cos n
n!
n→∞
35. an =
cos n
n2
37. an = (−1)n
36. an =
e−n
n
cos nπ
n2
38. an = (−1)n
2n + 1
n
4
14. an = √
n+1
In exercises 15–34, determine whether the sequence converges
or diverges.
5n − 1
2n 3 + 1
39. an =
n+3
n+2
40. an =
n−1
n+1
41. an =
en
n
42. an =
n
2n
43. an =
2n
(n + 1)!
44. an =
3n
(n + 2)!
45. an =
10n
n!
46. an =
n!
5n
3n + 1
2n 2 − 1
16. an =
17. an =
n2 + 1
n+1
18. an =
n2 + 1
n3 + 1
47. an =
3n 2 − 2
n2 + 1
48. an =
19. an =
n+2
3n − 1
20. an =
n+4
n+1
49. an =
sin(n 2 )
n+1
50. an = e1/n
15. an =
2
ln n
n2
In exercises 39–46, determine whether the sequence is increasing, decreasing or neither.
2
10. an = 2
n
n
n+1
27. an = ne−n
n→∞
In exercises 9–14, (a) find the limit of each sequence, (b) use the
definition to show that the sequence converges and (c) plot the
sequence on a calculator or CAS.
1
9. an = 3
n
26. an = sin πn
In exercises 35–38, use the Squeeze Theorem and Corollary 1.1
to prove that the sequence converges to 0 (given that
lim n1 = lim n12 = 0).
3
6. an =
n+4
8. an = (−1)n
25. an = cos πn
3
In exercises 47–50, show that the sequence is bounded.
6n − 1
n+3
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Section 8.1
51.
n
Numerically
the limits of the sequences an = 1 + n1
estimate
n
and bn = 1 − n1 . Compare the answers to e and e−1 .
52.
n
Numerically estimate the limits of the sequences an = 1 + n2
n
and bn = 1 − n2 . Compare the answers to e2 and e−2 .
53.
60.
n→∞
for any constant r. (Hint: Make the substitution n = rm.)
(n + 1)n
(n + 1)n+1
and
lim
.
n→∞
nn
nn
54.
Evaluate lim
55.
A packing company works with 12 square boxes. Show that for
n = 1, 2, 3, . . . , a total of n 2 disks of diameter 12
fit into a box.
n
2
Let an be the wasted area in a box with n disks. Compute an .
56.
The pattern of a sequence can’t always be determined from the
first few terms. Start with a circle, pick two points on the circle
and connect them with a line segment. The circle is divided
into a1 = 2 regions. Add a third point, connect all points and
show that there are now a2 = 4 regions. Add a fourth point,
connect all points and show that there are a3 = 8 regions. Is the
pattern clear? Show that a4 = 16 and then compute a5 for a
surprise!
57.
58.
n→∞
You have heard about the “population explosion.’’ The following dramatic warning is adapted from the article “Doomsday:
Friday 13 November 2026’’ by Foerster, Mora and Amiot in
Science (Nov. 1960). Start with a0 = 3.049 to indicate that the
world population in 1960 was approximately 3.049 billion.
Then compute a1 = a0 + 0.005a02.01 to estimate the population
in 1961. Compute a2 = a1 + 0.005a12.01 to estimate the population in 1962, then a3 = a2 + 0.005a22.01 for 1963, and so on.
Continue iterating and compare your calculations to the actual
populations in 1970 (3.721 billion), 1980 (4.473 billion) and
1990 (5.333 billion). Then project ahead to the year 2035.
Frightening, isn’t it?
Figure B
Figure C
Argue that the sides of the squares are determined by the
Fibonacci sequence of exercise 59.
√
n
n converges or
61.
Determine whether the sequence an =
diverges. (Hint: Use n 1/n = e(1/n) ln n .)
62.
Suppose that a1 = 1 and an+1 = 12 an + a4n . Show numerically that the sequence converges to 2. To find this limit analytically, let L = lim an+1 = lim an and solve the equation
n→∞
n→∞
L = 12 L + L4 .
63.
As in exercise 62, determine
the
limit of the sequence defined
by a1 = 1, an+1 = 12 an + acn for c > 0 and an > 0.
64.
Suppose that a ball is launched from the ground with
initial velocity v. Ignoring air resistance, it will rise to a
height of v 2 /(2g) and fall back to the ground at time t = 2v/g.
Depending on how “lively’’ the ball is, the next bounce will
only rise to a fraction of the previous height. The coefficient of
restitution r, defined as the ratio of landing velocity to rebound
velocity, measures the liveliness of the ball. The second bounce
has launch velocity rv, the third bounce has launch velocity
r 2 v and so on. It might seem that the ball will bounce forever.
To see that it does not, argue that the time to complete
(1 + r), the time to complete three
two bounces is a2 = 2v
g
(1 + r + r 2 ), etc. Take r = 0.5 and numerbounces is a3 = 2v
g
ically determine the limit of this sequence (we study this type
of sequence in detail in section 8.2). In particular, show that
(1 + 0.5) = 32 , (1 + 0.5 + 0.52 ) = 74 and (1 + 0.5 + 0.52 +
0.53 ) = 158 , find a general expression for an and determine the
limit of the sequence. Argue that at the end of this amount of
time, the ball has stopped bouncing.
65.
A surprising follow-up to the bouncing ball problem of
exercise 64 is found in An Experimental Approach to
Nonlinear Dynamics and Chaos by Tufillaro, Abbott and
Reilly. Suppose the ball is bouncing on a moving table that
oscillates up and down according to the equation A cos ωt for
The so-called hailstone sequence is defined by
3xk−1 + 1 if xk−1 is odd
.
xk = 1
x
if xk−1 is even
2 k−1
A different population model was studied by Fibonacci, an
Italian mathematician of the thirteenth century. He imagined a
population of rabbits starting with a pair of newborns. For one
month, they grow and mature. The second month, they have a
pair of newborn baby rabbits. We count the number of pairs of
rabbits. Thus far, a0 = 1, a1 = 1 and a2 = 2. The rules are:
adult rabbit pairs give birth to a pair of newborns every month,
newborns take one month to mature and no rabbits die. Show
In this exercise, we visualize the Fibonacci sequence. Start with
two squares of side 1 placed next to each other (see Figure A).
Place a square on the long side of the resulting rectangle (see
Figure B); this square has side 2. Continue placing squares on
the long sides of the rectangles: a square of side 3 is added in
Figure C, then a square of side 5 is added to the bottom of Figure C, and so on.
Figure A
If you start with x1 = 2n for some positive integer n, show that
xn+1 = 1. The question (an unsolved research problem) is
whether you eventually reach 1 from any starting value. Try
several odd values for x1 and show that you always reach 1.
59.
635
that a3 = 3, a4 = 5 and in general an = an−1 + an−2 . This
sequence of numbers, known as the Fibonacci sequence,
occurs in an amazing number of applications.
n
n
Given that lim 1 + n1 = e, show that lim 1 + nr = er
n→∞
Sequences of Real Numbers
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Chapter 8 Infinite Series
some amplitude A and frequency ω. Without the motion of the
table, the ball will quickly reach a height of 0 as in exercise 64.
For different values of A and ω, however, the ball can settle into
an amazing variety of patterns. To understand this, explain why
the collision between table and ball could subtract or add
velocity to the ball (what happens if the table is going up?
8.2
down?). A simplified model of the velocity of the ball at
successive collisions with the table is vn+1 = 0.8vn −
10 cos(v0 + v1 + · · · + vn ). Starting with v0 = 5, compute
v1 , v2 , . . . , v15 . In this case, the ball never settles into a pattern;
its motion is chaotic.
INFINITE SERIES
Recall that we write the decimal expansion of
1
3
as the repeating decimal
1
= 0.33333333̄,
3
where we understand that the 3’s in this expansion go on for ever and ever. An alternative
way to think of this is as
1
= 0.3 + 0.03 + 0.003 + 0.0003 + 0.00003 + · · ·
3
= 3(0.1) + 3(0.1)2 + 3(0.1)3 + 3(0.1)4 + · · · + 3(0.1)k + · · · .
(2.1)
For convenience, we write (2.1) using summation notation as
∞
1 =
[3(0.1)k ].
3
k=1
(2.2)
But, what exactly could we mean by the infinite sum indicated in (2.2)? Of course, you
can’t add infinitely many things together. (You can only add two things at a time.) By this
expression, we mean that as you add together more and more terms, the sum gets closer and
closer to 13 .
In general, for any sequence {ak }∞
k=1 , suppose we start adding the terms together. We
define the individual sums by
S1 = a1 ,
S2 = a1 + a2 = S1 + a2 ,
S3 = a1 + a2 + a3 = S2 + a3 ,
S2
S4 = a1 + a2 + a3 + a4 = S3 + a4 ,
S3
..
.
Sn = a1 + a2 + · · · + an−1 + an = Sn−1 + an
(2.3)
(2.4)
Sn−1
and so on. We refer to Sn as the nth partial sum. Note that we can compute any one of these
as the sum of two numbers: the nth term, an and the previous partial sum, Sn−1 , as indicated
in (2.4).
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Section 8.2
1
For instance, for the sequence k
2
∞
S1 =
S2 =
S3 =
S4 =
S5 =
Infinite Series
637
, consider the partial sums
k=1
1
,
2
1
1
3
+
= ,
2 22
4
3
1
7
+
= ,
4 23
8
7
1
15
+ 4 =
,
8 2
16
15
1
31
+ 5 =
16 2
32
1
3
and so on. Look at these carefully and you might notice that S2 = = 1 − 2 , S3 =
4
2
7 = 1 − 1 S = 15 = 1 − 1
1
, 4
and so on, so that Sn = 1 − n , for each n = 1, 2, . . . . If
3
4
8
16
2
2
2
we were to consider the convergence or divergence of the sequence {Sn }∞
n=1 of partial sums,
observe that we now have
1
lim Sn = lim 1 − n = 1.
n→∞
n→∞
2
about what this says: as we add together more and more terms of the sequence
Think
1 ∞
, the partial sums are drawing closer and closer to 1. In this instance, we write
2k k=1
∞
1
= 1.
(2.5)
k
2
k=1
It’s very important to understand what’s going on here. This new mathematical object,
∞ 1
is called a series (or infinite series). It is not a sum in the usual sense of the word,
k
k=1 2
but rather, the limit of the sequence of partial sums. Equation (2.5) says that as we add together more and more terms, the sums are approaching the limit of 1.
In general, for any sequence, {ak }∞
k=1 , we can write down the series
a1 + a2 + · · · + ak + · · · =
∞
ak .
k=1
n
ak converges (to some number S), then we say that
If the sequence of partial sums Sn =
∞
k=1
ak converges (to S). We write
the series
k=1
Definition of Infinite Series
∞
k=1
ak = lim
n→∞
n
k=1
ak = lim Sn = S.
n→∞
(2.6)
In this case, we call S the sum of the series. On the other hand, if the sequence of
lim Sn does not exist), then we say that the series
partial sums, {Sn }∞
n=1 diverges (i.e., n→∞
diverges.
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Chapter 8 Infinite Series
Example 2.1
A Convergent Series
Determine if the series
Solution
∞ 1
converges or diverges.
k
k=1 2
From our work on the introductory example, observe that
n
∞
1
1
= lim
ak = lim 1 − n = 1.
n→∞
n→∞
2k
2
k=1
k=1
In this case, we say that the series converges to 1.
■
In the following example, we examine a simple divergent series.
Example 2.2
A Divergent Series
∞
Investigate the convergence or divergence of the series
k2.
k=1
Solution
Here, we have the nth partial sum
n
Sn =
k 2 = 12 + 22 + · · · + n 2
k=1
and
lim Sn = lim (12 + 22 + · · · + n 2 ) = ∞.
Sn
n→∞
n→∞
Since the sequence of partial sums diverges, the series diverges also.
1.0
■
0.8
Determining the convergence or divergence of a series is only rarely as simple as it
was in examples 2.1 and 2.2.
0.6
0.4
Example 2.3
0.2
n
5
10
15
20
Figure 8.17
n
1
.
Sn =
k=1 k(k + 1)
Sn =
n
n
k=1
1
k(k + 1)
10
0.90909091
100
0.99009901
1000
0.999001
10,000
0.99990001
100,000
0.99999
1 × 106
0.999999
1 × 10
0.9999999
7
A Series with a Simple Expression for the Partial Sums
Investigate the convergence or divergence of the series
∞
1
.
k(k
+ 1)
k=1
Solution
In Figure 8.17, we have plotted the first 20 partial sums. In the accompanying table, we list a number of partial sums of the series.
From both the graph and the table, it appears that the partial sums are approaching 1,
as n → ∞. However, we must urge caution. It is extremely difficult to look at a graph or
a table of any finite number of the partial sums and decide whether a given series is converging or diverging. In the present case, we are fortunate that we can find a simple
expression for the partial sums. We leave it as an exercise to find the partial fractions
decomposition of the general term of the series:
1
1
1
= −
.
k(k + 1)
k
k+1
(2.7)
Now, consider the nth partial sum. From (2.7), we have
n
n 1
1
1
Sn =
=
−
k(k + 1)
k
k+1
k=1
k=1
1 1
1 1
1 1
1
1
1
1
=
−
+
−
+
−
+ ··· +
−
+
−
.
1 2
2 3
3 4
n−1 n
n
n+1
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Section 8.2
639
Infinite Series
Notice how nearly every term in the partial sum is canceled by another term in the sum
(the next term). For this reason, such a sum is referred to as a telescoping (or collapsing) sum. We now have
Sn = 1 −
and so,
1
n+1
lim Sn = lim 1 −
n→∞
n→∞
1
n+1
= 1.
This says that the series converges to 1, as conjectured from the graph and the table.
■
It is relatively rare that we know the sum of a convergent series exactly. Usually, we
must test a series for convergence using some indirect method and then approximate the
sum by calculating some partial sums. We now consider one additional type of series
∞ 1
whose sum is known exactly. The series we considered in example 2.1,
, is an
k
k=1 2
example of a type of series called geometric series. We have the following result.
Theorem 2.1
∞
ar k converges to
For a = 0, the geometric series
k=0
|r| ≥ 1. (Here, r is referred to as the ratio.)
a
if |r| < 1 and diverges if
1−r
Proof
The proof relies on a clever observation. Note that the first term of the series corresponds to k = 0 and so, the nth partial sum (the sum of the first n terms) is
Sn = a + ar 1 + ar 2 + · · · + ar n−1 .
(2.8)
Multiplying (2.8) by r, we get
r Sn = ar 1 + ar 2 + ar 3 + · · · + ar n .
(2.9)
Subtracting (2.9) from (2.8), we get
(1 − r)Sn = (a + ar 1 + ar 2 + · · · + ar n−1 ) − (ar 1 + ar 2 + ar 3 + · · · + ar n )
= a − ar n = a(1 − r n ).
Dividing both sides by (1 − r) gives us
Sn =
a(1 − r n )
.
1−r
If |r| < 1, notice that r n → 0 as n → ∞ and so,
a
a(1 − r n )
=
.
n→∞
1−r
1−r
lim Sn = lim
n→∞
We leave it as an exercise to show that if |r| ≥ 1, lim Sn does not exist.
n→∞
■
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Chapter 8 Infinite Series
Example 2.4
A Convergent Geometric Series
k
∞
1
5
.
Investigate the convergence or divergence of the series
3
k=2
Solution
The first 20 partial sums are plotted in Figure 8.18. It appears from the
graph that the sequence of partial sums is converging to some number around 0.8. Further evidence is found in the following table, showing the first 20 partial sums.
Sn
n
1.0
Sn =
n+1
k=2
0.8
0.6
0.4
0.2
n
5
10
15
20
Figure 8.18
k
n+1
1
5·
.
Sn =
3
k=2
5
k
1
3
n
Sn =
n+1
5
k=2
k
1
3
1
0.55555556
11
0.83332863
2
0.74074074
12
0.83333177
3
0.80246914
13
0.83333281
4
0.82304527
14
0.83333316
5
0.82990398
15
0.83333328
6
0.83219021
16
0.83333331
7
0.83295229
17
0.83333333
8
0.83320632
18
0.83333333
9
0.833291
19
0.83333333
0.83331922
20
0.83333333
10
The table suggests that the series converges to approximately 0.83333333. Again, we
must urge caution. Graphical and numerical evidence can be very misleading when
examining series. Some sequences and series converge (or diverge) far too slowly to
observe graphically or numerically. You must always confirm your suspicions with
careful mathematical analysis. In the present case, note that while the series is not quite
written in the usual form, it is a geometric series, as follows:
k
2
3
4
n
∞
1
1
1
1
1
5
=5
+5
+5
+ ··· + 5
+ ···
3
3
3
3
3
k=2
2 2
1
1
1
=5
1+ +
+ ···
3
3
3
∞
1 2 1 k
=
5
.
3
3
k=0
2
You can now see that this is a geometric series with ratio r = 13 and a = 5 13 . Further,
since
1
|r| = < 1,
3
we have from Theorem 2.1 that the series converges to
5
2
5 13
a
5
1 = 92 = = 0.83333333̄,
=
1−r
6
1− 3
3
which is consistent with the graph and the table of partial sums.
■
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Section 8.2
Example 2.5
Infinite Series
641
A Divergent Geometric Series
Investigate the convergence or divergence of the series
7 k
6 −
.
2
k=0
∞
Solution
A graph showing the first 20 partial sums (see Figure 8.19) is not particularly helpful, until you look at the vertical scale. The following table showing the values of the first 20 partial sums is more revealing.
Sn
n
0.5 1011
Sn =
7 k
6 −
2
k=0
n−1
n
Sn =
7 k
6 −
2
k=0
n−1
1
6
11
1.29 × 106
2
−15
12
−4.5 × 106
0.5 1011
3
58.5
13
1.6 × 107
1.0 1011
4
−198.75
14
−5.5 × 107
1.5 1011
5
701.63
15
1.9 × 108
6
−2449.7
16
−6.8 × 108
7
8579.9
17
2.4 × 109
8
−30,024
18
−8.3 × 109
9
1.05 × 105
19
2.9 × 1010
−3.68 × 105
20
−1 × 1011
n
5
10
15
20
Figure 8.19
k
n−1
7
6· −
.
Sn =
2
k=0
10
Note that while the partial sums are oscillating back and forth between positive and negative values, they are growing larger and larger in absolute value. We can confirm our
suspicions by observing that this is a geometric series with ratio r = − 72 . Since
7 7
|r| = − = ≥ 1,
2
2
the series is divergent, as we suspected.
■
You will find that determining whether a series is convergent or divergent usually
involves a lot of hard work. The following simple observation provides us with a very
useful test.
Theorem 2.2
∞
ak converges, then lim ak = 0.
If
k→∞
k=1
Proof
∞
ak converges to some number L. This means that the sequence of
n
ak also converges to L. Notice that
partial sums defined by Sn =
Suppose that
k=1
k=1
Sn =
n
k=1
ak =
n−1
k=1
ak + an = Sn−1 + an .
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Chapter 8 Infinite Series
Subtracting Sn−1 from both sides, we have
an = Sn − Sn−1 .
This gives us
lim an = lim (Sn − Sn−1 ) = lim Sn − lim Sn−1 = L − L = 0,
n→∞
n→∞
n→∞
n→∞
as desired.
■
The following very useful test follows directly from Theorem 2.2.
kth-Term Test for Divergence
If lim ak = 0, then the series
Sn
k→∞
∞
ak diverges.
k=1
20
15
The kth-term test is so simple, you should use it to test every series you run into. It says that
if the terms don’t tend to zero, the series is divergent and there’s nothing more to do. However, as we’ll soon see, if the terms do tend to zero, there is no guarantee that the series
converges and additional testing is needed.
10
5
n
5
10
15
Example 2.6
A Series Whose Terms Do Not Tend to Zero
20
Figure 8.20
n
k
Sn =
.
k
+
1
k=1
Investigate the convergence or divergence of the series
∞
k
.
k
+
1
k=1
Solution
A graph showing the first 20 partial sums is shown in Figure 8.20. The
partial sums appear to be increasing without bound as n increases. A table of values
would indicate the same sort of growth. (Try this!) We can resolve the question of convergence quickly by observing that
lim
k→∞
k
= 1 = 0.
k+1
From the kth-term test for divergence, the series must diverge.
■
This next example shows an important series whose terms tend to 0 as k → ∞, but
that diverges, nonetheless.
CAUTION
The converse of Theorem 2.2 is
false. That is, having lim a k = 0
k→∞
Example 2.7
The Harmonic Series
Investigate the convergence or divergence of the harmonic series:
∞ 1
.
k=1 k
does not guarantee that the series
∞
a k converges. Be very clear
k=1
about this point. This is a very
common misconception.
Solution
In Figure 8.21, we see the first 20 partial sums of the series. In the following table, we display the first 20 partial sums. (You could do the same with the first
200 or 2000 partial sums and there would be little difference in your conclusion.) The
table and the graph suggest that the series might converge to a number around 3.6. As
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Section 8.2
Sn
Sn =
2
1
n
5
10
15
Figure 8.21
n 1
Sn =
.
k=1 k
20
Sn =
1
1
11
3.01988
2
1.5
12
3.10321
3
1.83333
13
3.18013
4
2.08333
14
3.25156
5
2.28333
15
3.31823
6
2.45
16
3.38073
7
2.59286
17
3.43955
8
2.71786
18
3.49511
9
2.82897
19
3.54774
10
2.92897
20
3.59774
n
643
n 1
k=1 k
n
4
3
n 1
k=1 k
Infinite Series
always with sequences and series, we need to confirm this suspicion. From our test for
divergence, we have
1
lim ak = lim = 0.
k→∞
k→∞ k
Be careful: once again, this does not say that the series converges. If the limit had been
nonzero, we would have stopped and concluded that the series diverges. In the present
case, where the limit is 0, we can only conclude that further study is needed. (That is,
the series may converge, but we will need to investigate further.)
The following clever proof provides a preview of things to come. Consider the nth
partial sum
n
1
1 1 1
1
Sn =
= + + + ··· + .
k
1 2 3
n
k=1
Note that Sn corresponds to the sum of the areas of the n rectangles superimposed on the
graph of y = x1 shown in Figure 8.22 for the case where n = 7.
y
1.2
1.0
0.8
0.6
0.4
0.2
x
1
2
3
4
5
Figure 8.22
1
y= .
x
6
7
8
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Chapter 8 Infinite Series
Notice that since each of the indicated rectangles lies partly above the curve, we have
Sn = Sum of areas of n rectangles
n+1
1
dx
≥ Area under the curve =
x
1
n+1
= ln |x|
= ln(n + 1).
(2.10)
1
However, the sequence {ln(n + 1)}∞
n=1 diverges, as
lim ln(n + 1) = ∞.
n→∞
Since Sn ≥ ln(n + 1), for all n [from (2.10)], we must also have that lim Sn = ∞.
n→∞
∞ 1
From the definition of convergence of a series, we now have that
diverges, too,
k=1 k
1
even though lim = 0.
k→∞ k
■
We complete this section with several unsurprising results.
Theorem 2.3
∞
∞
∞
ak converges to A and
bk converges to B, then the series
(ak ± bk )
(i) If
k=1
converges to A ± B and
(ii) If
∞
ak converges and
k=1
∞
k=1
k=1
(c ak ) converges to c A, for any constant, c.
k=1
∞
∞
k=1
k=1
bk diverges, then
(ak ± bk ) diverges.
The proof of the theorem is left as an exercise.
EXERCISES 8.2
1.
Suppose that your friend is confused about the difference
between the convergence of a sequence and the convergence of a series. Carefully explain the difference between
k
and
convergence or divergence of the sequence ak =
k+1
∞
k
the series
.
k=1 k + 1
2.
∞
hold if the series starts at k = 2, or k = 3 or at any positive
integer.
4.
Explain in words why the kth term test for divergence is
true. Explain why it is not true that if lim ak = 0 then
k→∞
ak necessarily converges. In your explanation, include
k=1
an important example that proves that this is not true and comment on the fact that the convergence of ak to 0 can be slow or
fast.
3.
In Theorems 2.2 and 2.3, the series starts at k = 1, as in
∞
ak . Explain why the conclusions of the theorems
k=1
We emphasized in the text that numerical and graphical
evidence for the convergence of a series can be misleading. Suppose your calculator carries 14 digits in its
calculations. Explain why for large enough values of n, the
n 1
term n1 will be too small to change the partial sum
. Thus,
k=1 k
the calculator would incorrectly indicate that the harmonic
series converges.
In exercises 5–28, determine if the series converges or diverges.
For convergent series, find the sum of the series.
5.
k
∞
1
3
5
k=0
6.
∞
1
k=0
3
(5)k
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Section 8.2
∞
1 k
1
−
7.
2
3
k=0
9.
∞
1
k=0
11.
∞
k=1
13.
15.
k=1
19.
(3)k
4
k(k + 2)
12.
14.
∞
16.
k
2k + 1
k 2 (k + 1)2
18.
3−k
20.
∞ 2
1
+ k
23.
3k
2
k=0
25.
(−1)k
k=0
27.
∞
2e−k
k=0
22.
1
− k
k
4
k=1
3
2k
(−1)k+1
k=0
24.
∞ 1
1
−
2k
3k
k=0
∞ 1
26.
∞
3k
k+1
28.
∞
Compute several partial sums of the series 1 + 1 − 1 +
1 − 1 + 1 − 1 + · · · . Argue that the limit of the sequence of
partial sums does not exist, so that the series diverges. Also,
write this series as a geometric series and use Theorem 2.1 to
conclude that the series diverges. Finally, use the kth-term test
for divergence to conclude that the series diverges.
41.
Write 0.99999̄ = 0.9 + 0.09 + 0.009 + · · · and sum the geometric series to prove that 0.99999̄ = 1.
42.
As in exercise 41, prove that 0.199999̄ = 0.2.
43.
Suppose you have n boards of length L. Place the first board
L
with length 2n
hanging over the edge of the table. Place the
L
next board with length 2(n−1)
hanging over the edge of the first
L
board. The next board should hang 2(n−2)
over the edge of the
second board. Continuing on until the last board hangs L2 over
the edge of the (n − 1)st board. Theoretically, this stack will
balance (in practice, don’t use quite as much overhang). With
n = 8, compute the total overhang of the stack. Determine the
number of boards n such that the total overhang is greater than
L. This means that the last board is entirely beyond the edge of
the table. What is the limit of the total overhang as n → ∞?
(−1)k+1
4
3k
(−1)k
k3
+1
k=0
k=0
k2
In exercises 29–34, use graphical and numerical evidence to
conjecture the convergence or divergence of the series.
∞
∞
∞
1
1
3
√
29.
31.
30.
2
k
k!
k
k=1
k=1
k=1
32.
∞
2k
k=1
35.
33.
k
Prove that if
∞
4k
k=1
∞
34.
k2
ak converges, then
44.
ak converges for any
∞
ak converges to L, what
positive integer m. In particular, if
∞
k=1
ak converge to?
does
k=m
36.
Prove that if
∞
ak diverges, then
∞
ak diverges for any
k=m
k=1
positive integer m.
38.
...
L
4
k!
k=m
k=1
L
2
∞
2k
k=1
∞
1
3
1
5
40.
4
k+1
∞
−k
e − e−(k+1)
∞
> 14 , we have
n 1
.
k=1 k
+ 14 >
+ 14 = 12 . Therefore, S4 > 32 + 12 = 2. Similarly, + 16 +
+ 18 > 18 + 18 + 18 + 18 = 12 , so S8 > 52 . Show that S16 > 3
and S32 > 72 . For which n can you guarantee that Sn > 4?
Sn > 5? For any positive integer m, determine n such that
Sn > m. Conclude that the harmonic series diverges.
9
k(k + 3)
∞
1
3
1
4
1
7
k=0
∞ 1
1
−
2k
k+1
k=0
∞
∞
k=0
k=0
21.
Show that S1 = 1 and S2 = 32 . Since
∞
4k
k+2
k=1
k=1
∞
2
∞
another proof of the divergence of this series. Let Sn =
∞
1 k
5 −
10.
3
k=0
∞
3k
k
+4
k=1
k=1
17.
2
k
∞
1
4
8.
2
k=0
645
Infinite Series
37.
Prove Theorem 2.3 (i).
Prove Theorem 2.3 (ii).
39.
The harmonic series is probably the single most important
series to understand. In this exercise, we guide you through
Have you ever felt that the line you’re standing in moves more
slowly than the other lines? In An Introduction to Probability
Theory and Its Applications, William Feller proved just how
bad your luck is. Let N be the number of people who get in line
until someone waits longer than you do (you’re the first, so
N ≥ 2). The probability that N = k is given by p(k) =
1
. Prove that the total probability equals 1; that is,
k(k − 1)
∞
1
= 1. From probability theory, the average (mean)
k(k
− 1)
k=2
number of people who must get in line before someone has
∞
1
k
. Prove that
waited longer than you is given by
k=2 k(k − 1)
this diverges to ∞. Talk about bad luck!
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45.
46.
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Chapter 8 Infinite Series
If 0 < r < 12 , show that 1 + 2r + 4r 2 + · · · + (2r)n + · · · =
1
1
. Replace r with
and discuss what’s interesting
1 − 2r
1000
500
about the decimal representation of
.
499
In exercise 64 of section 8.1, you showed that a particular
bouncing ball takes 2 seconds to complete its infinite number of
bounces. In general, the total time it takes for a ball to complete
∞
2v r k and the total distance the ball moves is
its bounces is
g
k=0
2 v ∞ 2k
r . Assuming 0 < r < 1, find the sums of these geog k=0
metric series.
47.
To win a deuce tennis game, one player or the other must win the
next two points. If each player wins one point, the deuce starts
over. If you win each point with probability p, the probability
that you win the next two points is p2 . The probability that you
win one of the next two points is 2 p(1 − p). The probability that
you win a deuce game is then p2 + 2 p(1 − p) p2 +
[2 p(1 − p)]2 p2 + [2 p(1 − p)]3 p2 + · · · . Explain what each
term represents, explain why the geometric series converges and
find the sum of the series. If p = 0.6, you’re a better player than
your opponent. Show that you are more likely to win a deuce
game than you are a single point. The slightly strange scoring
rules in tennis make it more likely that the better player wins.
48.
On an analog clock, at 1:00 the minute hand points to 12 and
the hour hand points to 1. When the minute hand reaches 1, the
hour hand has progressed to 1 + 121 . When the minute hand
reaches 1 + 121 , the hour hand has moved to 1 + 121 + 121 2 . Find
the sum of a geometric series to determine the time at which the
minute hand and hour hand are in the same location.
49.
50.
Two bicyclists are 40 miles apart, riding toward each other at
20 mph (each). A fly starts at one bicyclist and flies toward the
other bicyclist at 60 mph. When it reaches the bike, it turns
around and flies back to the first bike. It continues flying back
and forth until the bikes meet. Determine the distance flown on
each leg of the fly’s journey and find the sum of the geometric
series to get the total distance flown. Verify that this is the right
answer by solving the problem the easy way.
Give an example where
∞
(ak + bk ) converges.
∞
k=1
ak and
∞
reduction in the total amount of counterfeit money successfully
spent.
53.
A dosage d of a drug is given at times t = 0, 1, 2, . . . . The drug
decays exponentially with rate r in the bloodstream. The
amount in the bloodstream after n + 1 doses is d + de−r +
de−2r + · · · + de−nr . Show that the eventual level of the drug
d
. If r = 0.1,
(after an “infinite’’ number of doses) is
1 − e−r
find the dosage needed to maintain a drug level of 2.
54.
The Cantor set is one of the most famous sets in mathematics.
To construct the Cantor set, start
with
the interval [0, 1]. Then
the middle third, 13 , 23 . This leaves the set
remove
0, 13 ∪ 23 , 1 . For each of the two subintervals, remove
1 2 the
,
and
middle
third;
in
this
case,
remove
the
intervals
9 9
7 8
,
.
Continue
in
this
way,
removing
the
middle
thirds
of
9 9
each remaining interval. The Cantor set is all points in [0, 1]
that are not removed. Argue that 0, 1, 13 and 23 are in the Cantor
set, and identify four more points in the set. It can be shown
that there are an infinite number of points in the Cantor set. On
the other
hand, the total length of the subintervals removed is
1
+ 2 19 + · · · . Find the third term in this series, identify the
3
series as a convergent geometric series and find the sum of the
series. Given that you started with an interval of length 1, how
much “length” does the Cantor set have?
55.
In this exercise, we will find the present value of a plot of
farmland. Assume that a crop of value $c will be planted in
years 1, 2, 3, and so on, and the yearly inflation rate is r. The
present value is given by
P = ce−r + ce−2r + ce−3r + · · · .
Find the sum of the geometric series to compute the present value.
56.
p=prime
taken over the prime numbers, not all integers. Compare
your results to the number 62 .
π
bk both diverge but
k=1
k=1
51.
Suppose $100,000 of counterfeit money is introduced into the
economy. Each time the money is used, 25% of the remaining
money is identified as counterfeit and removed from circulation. Determine the total amount of counterfeit money successfully used in transactions. This is an example of the multiplier
effect in economics.
52.
In exercise 51, suppose that a new marking scheme on dollar
bills helps raise the detection rate to 40%. Determine the
Infinite products are also of great interest to mathematicians. Numerically explore
or
theconvergence
divergence of the infinite product 1 − 14 1 − 19 1 − 251
1 − p12 . Note that the product is
1 − 491 · · · =
57.
In example 2.7, we showed that 1 + 12 + 13 + · · · +
1
1
> ln(n + 1). Superimpose the graph of f (x) = x −1
n
1
1
1
onto Figure 8.22 and show that 2 + 3 + · · · + n < ln(n).
Conclude that ln(n + 1) < 1 + 12 + 13 + · · · + n1 < 1 + ln(n).
Euler’s constant is defined by
1 1
1
γ = lim 1 + + + · · · + − ln(n) .
n→∞
2 3
n
Look up the value of γ . (Hint: Use your CAS.) Use γ to estin 1
for n = 10,000 and n = 100,000.
mate
i =1 i
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Section 8.3
8.3
The Integral Test and Comparison Tests
647
THE INTEGRAL TEST AND COMPARISON TESTS
As we have observed a number of times now, we are usually unable to determine the sum of
a convergent series. In fact, for most series we cannot determine whether they converge or
diverge by simply looking at the sequence of partial sums. Most of the time, we will need to
test a series for convergence in some indirect way. If we find that the series is convergent, we
can then approximate its sum by numerically computing some partial sums. In this section,
we will develop additional tests for convergence of series. The first of these is a generalization of the method we used to show that the harmonic series was divergent in section 8.2.
∞
ak , suppose that there is a function f for which
For a given series
k=1
f (k) = ak , for k = 1, 2, . . . ,
where f is continuous, decreasing and f (x) ≥ 0 for all x ≥ 1. We consider the nth partial
sum
n
Sn =
ak = a1 + a2 + · · · + an .
k=1
Look carefully at Figure 8.23a. We have constructed (n − 1) rectangles on the interval
[1, n], each of width 1 and with height equal to the value of the function at the right-hand
endpoint of the subinterval on which the rectangle is constructed. Notice that since each
rectangle lies completely beneath the curve, the sum of the areas of the (n − 1) rectangles
shown is less than the area under the curve from x = 1 to x = n. That is,
n
f (x) dx. (3.1)
0 ≤ Sum of areas of (n − 1) rectangles ≤ Area under the curve =
y
y f (x)
(2, a2)
(3, a3)
(n, an)
1
x
1
2
3
4
n
Figure 8.23a
(n − 1) rectangles, lying beneath
the curve.
Note that the area of the first rectangle is length × width = (1)(a2 ), the area of the second
rectangle is (1)(a3 ) and so on. We get that the sum of the areas of the (n − 1) rectangles
indicated in Figure 8.23a is
a2 + a3 + a4 + · · · + an = Sn − a1 ,
since
Sn = a1 + a2 + · · · + an .
Together with (3.1), this gives us
0 ≤ Sum of areas of (n − 1) rectangles
= Sn − a1 ≤ Area under the curve =
Now, suppose that the improper integral
have
y
y f (x)
(1, a1)
n
0 ≤ Sn − a1 ≤
∞
1
(n 1, an1)
(3.2)
f (x) dx converges. Then, from (3.2), we
f (x) dx ≤
Adding a1 to all the terms gives us
f (x) dx.
1
1
(2, a2)
n
∞
f (x) dx.
1
a1 ≤ Sn ≤ a1 +
∞
f (x) dx.
1
x
1
2
3
4
n
Figure 8.23b
(n − 1) rectangles, partially above
the curve.
∞
This says that the sequence of partial sums {Sn }∞
n=1 is bounded. Since {Sn }n=1 is also
∞
ak is
monotonic (why is that?), {Sn }∞
n=1 is convergent by Theorem 1.4 and so, the series
k=1
also convergent.
In Figure 8.23b, we have constructed (n − 1) rectangles on the interval [1, n], each of
width 1, but with height equal to the value of the function at the left-hand endpoint of the
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Chapter 8 Infinite Series
HISTORICAL NOTES
subinterval on which the rectangle is constructed. In this case, the sum of the areas of the
(n − 1) rectangles shown is greater than the area under the curve. That is,
n
0 ≤ Area under the curve =
f (x) dx
1
≤ Sum of areas of (n − 1) rectangles.
(3.3)
Further, note that the area of the first rectangle is length × width = (1)(a1 ), the area of the
second rectangle is (1)(a2 ) and so on. We get that the sum of the areas of the (n − 1)
rectangles indicated in Figure 8.23b is
Colin Maclaurin (1698–1746)
Scottish mathematician who
discovered the Integral Test.
Maclaurin was one of the founders
of the Royal Society of Edinburgh
and was a pioneer in the
mathematics of actuarial studies.
The Integral Test was introduced in
a highly influential book that also
included a new treatment of an
important method for finding series
of functions. Maclaurin series, as
we now call them, are developed in
section 8.7.
a1 + a2 + · · · + an−1 = Sn−1 .
Together with (3.3), this gives us
0 ≤ Area under the curve =
n
f (x) dx
1
≤ Sum of areas of (n − 1) rectangles = Sn−1 .
(3.4)
∞
Now, suppose
n that the improper integral 1 f (x) dx diverges. Since f (x) ≥ 0, this says
that lim 1 f (x) dx = ∞. From (3.4), we have that
n→∞
n
f (x) dx ≤ Sn−1 .
1
This says that
lim Sn−1 = ∞,
n→∞
also. So, the sequence of partial sums {Sn }∞
n=1 diverges and hence, the series
diverges, too.
We summarize the results of this analysis in the following theorem.
∞
ak
k=1
Theorem 3.1 (Integral Test)
If f (k) = ak for all k = 1, 2, . . . and f is continuous, decreasing and f (x) ≥ 0, for
∞
∞
ak either both converge or both diverge.
x ≥ 1, then 1 f (x) dx and
k=1
It is important to recognize that while the Integral Test might say that a given series
and improper integral both converge, it does not say that they will converge to the same
thing. In fact, this is generally not true, as we see in the following example.
Sn
2.5
2.0
Example 3.1
1.5
Using the Integral Test
Investigate the convergence or divergence of the series
∞
k=0
1.0
0.5
n
5
10
15
Figure 8.24
n−1
1
Sn =
.
2
k=0 k + 1
20
k2
1
.
+1
Solution
The graph of the first 20 partial sums shown in Figure 8.24 suggests that
the series converges to some value around 2. In the accompanying table (found in the
margin on the following page), you can find some selected partial sums. We have shown
so many partial sums due to the very slow rate of convergence of this series.
Based on our computations, we cannot say whether the series is converging very
slowly to a limit around 2.076 or whether the series is instead diverging very slowly,
as we saw earlier for the harmonic series. To determine which is the case, we must test
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Section 8.3
n
Sn =
n−1
k=0
1
k2 + 1
10
1.97189
50
2.05648
100
2.06662
200
2.07166
500
2.07467
1000
2.07567
2000
2.07617
The Integral Test and Comparison Tests
649
1 .
the series further. Define f (x) = 2
Note that f is continuous and positive everyx +1
1
= ak , for all k ≥ 1. Further,
where and f (k) = 2
k +1
f (x) = (−1)(x 2 + 1)−2 (2x) < 0,
for x ∈ (0, ∞), and so, f is decreasing. This says that the Integral Test applies to this
series. So, we consider the improper integral
R
∞
R
1
1
−1 dx = lim
dx = lim tan x 2
2
R→∞ 0 x + 1
R→∞
x +1
0
0
π
π
= lim (tan−1 R − tan−1 0) = − 0 = .
R→∞
2
2
By the Integral Test, we have that since the improper integral converges, the series must
converge, also. Since we have now established that the series is convergent, we can use
our earlier calculations to arrive at the estimated sum 2.076. Notice that this is not the
π
same as the value of the corresponding improper integral, which is ≈ 1.5708.
2
■
In the following example, we discuss an important type of series.
Example 3.2
The p-Series
Determine for which values of p the series
∞ 1
(a p-series) converges.
p
k=1 k
First, notice that for p = 1, this is the harmonic series, which diverges.
1
For p > 1, define f (x) = p = x − p . Notice that for x ≥ 1, f is continuous and
x
positive. Further,
Solution
f (x) = − px − p−1 < 0,
so that f is decreasing for x ≥ 1. This says that the Integral Test applies. We now consider
∞
R
x − p+1 R
x − p dx = lim
x − p dx = lim
R→∞ 1
R→∞ − p + 1 1
1
− p+1
Since p > 1 implies
R
1
1
= lim
−
=
.
that − p + 1 < 0.
R→∞ − p + 1
−p + 1
−p + 1
In this case, the improper integral converges and so too, must the series. In the case
where p < 1, we leave it as an exercise to show that the series diverges.
■
We summarize the result of example 3.2 as follows.
p-series
The p-series
∞ 1
converges if p > 1 and diverges if p ≤ 1.
p
k=1 k
Notice that in examples 3.1 and 3.2, we were able to use the Integral Test to establish
the convergence of several series. So, now what? We have observed that you can use the
partial sums of a series to estimate its value, but just how precise is a given estimate? We
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Chapter 8 Infinite Series
answer this question with the following result. First, if we estimate the sum s of the series
∞
n
ak by the nth partial sum Sn =
ak , we define the remainder Rn to be
y
k=1
y f (x)
Rn = s − Sn =
x
n1
Figure 8.25
Estimate of the remainder.
∞
k=1
(n 1, an1)
n
k=1
ak −
n
k=1
ak =
∞
ak .
k=n+1
Notice that this says that the remainder is the error in approximating s by Sn. For any series
shown to be convergent by the Integral Test, we can estimate the size of the remainder, as
follows. From Figure 8.25, observe that the remainder Rn corresponds to the sum of the
areas of the indicated rectangles. Further, under the conditions of the IntegralTest, this is
∞
less than the area under the curve y = f (x). (Recall that this area is finite, as 1 f (x) dx
converges.) That is, we have the following result.
Theorem 3.2 (Error Estimate for the Integral Test)
Suppose that f (k) = ak for all k = 1, 2, . . . , where
∞ f is continuous, decreasing and
f (x) ≥ 0 for all x ≥ 1. Further, suppose that 1 f (x) dx converges. Then, the remainder Rn satisfies
∞
∞
0 ≤ Rn =
ak ≤
f (x) dx.
k=n+1
n
We can use Theorem 3.2 to estimate the error in using a partial sum to approximate the
sum of a series.
Example 3.3
Estimating the Error in a Partial Sum
Estimate the error in using the partial sum S100 to approximate the sum of the series
∞ 1
.
3
k=1 k
Solution
First, recall that in example 3.2, we had shown that this series (a p-series,
with p = 3) is convergent, by the Integral Test. From Theorem 3.2 then, the remainder
term satisfies
R
∞
1 R
1
1
0 ≤ R100 ≤
dx = lim
dx = lim − 2
3
R→∞ 100 x 3
R→∞
2x 100
100 x
−1
1
= lim
+
= 5 × 10−5 .
R→∞ 2R 2
2(100)2
■
A more interesting and far more practical question related to example 3.3 is to use
Theorem 3.2 to help us determine the number of terms of the series necessary to obtain a
given accuracy.
Example 3.4
Finding the Number of Terms Needed
for a Given Accuracy
Determine the number of terms needed to obtain an approximation to the sum of the
∞ 1
series
correct to within 10−5 .
3
k=1 k
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Section 8.3
The Integral Test and Comparison Tests
651
Solution
Again, we already used the Integral Test to show that the series in question converges. Then, by Theorem 3.2, we have that the remainder satisfies
∞
R
1
1
1 R
dx
=
lim
dx
=
lim
−
0 ≤ Rn ≤
R→∞ n x 3
R→∞
x3
2x 2 n
n
−1
1
1
= lim
+ 2 = 2.
R→∞ 2R 2
2n
2n
So, to ensure that the remainder is less than 10−5 , we require
0 ≤ Rn ≤
Solving this last inequality for n yields
105
n ≥
2
2
or n ≥
1
≤ 10−5 .
2n 2
√
105
= 100 5 ≈ 223.6.
2
So, taking n ≥ 224 will guarantee the required accuracy and consequently, we have
∞ 1
224
1
≈
= 1.202047, which is correct to within 10−5 , as desired.
3
3
k
k
k=1
k=1
■
Comparison Tests
We next present two results that will allow us to compare a given series with one that is already known to be convergent or divergent, much as we did with improper integrals in section 7.7.
Theorem 3.3 (Comparison Test)
Suppose that 0 ≤ ak ≤ bk , for all k.
∞
∞
bk converges, then
ak converges, too.
(i) If
(ii)
If
k=1
∞
k=1
k=1
ak diverges, then
∞
bk diverges, too.
k=1
Intuitively, this theorem should make abundant sense: if the “larger” series converges,
then the “smaller” one must also converge. Likewise, if the “smaller” series diverges, then
the “larger” one must diverge, too.
Proof
Given that 0 ≤ ak ≤ bk for all k, observe that the nth partial sums of the two series
satisfy
(i) If
∞
0 ≤ Sn = a1 + a2 + · · · + an ≤ b1 + b2 + · · · + bn .
bk converges (say to B), this says that
k=1
0 ≤ Sn ≤ a1 + a2 + · · · + an ≤ b1 + b2 + · · · + bn ≤
∞
k=1
bk = B,
(3.5)
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Chapter 8 Infinite Series
for all n ≥ 1. From (3.5), the sequence {Sn }∞
n=1 of partial sums of
∞
ak is bounded.
k=1
Notice that {Sn }∞
n=1 is also increasing. (Why?) Since every bounded, monotonic se∞
ak is convergent, too.
quence is convergent (see Theorem 1.4), we get that
k=1
∞
ak is divergent, we have (since all of the terms of the series are nonnegative)
(ii) If
k=1
that
lim (b1 + b2 + · · · + bn ) ≥ lim (a1 + a2 + · · · + an ) = ∞.
n→∞
Thus,
∞
n→∞
bk must be divergent, also.
k=1
■
You can use the Comparison Test to test the convergence of series that look similar to series that you already know are convergent or divergent (notably, geometric series or p-series).
Example 3.5
Using the Comparison Test for a Convergent Series
Investigate the convergence or divergence of
∞
k=1
1
.
k 3 + 5k
Solution
From the graph of the first 20 partial sums shown in Figure 8.26, it
appears that the series converges to some value near 0.3. To confirm such a conjecture,
we must carefully test the series. Note that for large values of k, the general term of the
1
series looks like 3 , since when k is large, k 3 is much larger than 5k. This observation
k
∞ 1
is significant, since we already know that
is a convergent p-series ( p = 3 > 1).
3
k=1 k
Further, observe that
Sn
0.3
0.2
0.1
k3
n
5
10
15
20
Figure 8.26
n
1
.
Sn =
3
k=1 k + 5k
Sn
1
1
≤ 3,
+ 5k
k
∞ 1
∞
1
converges,
the
Comparison
Test
says
that
3
3
k=1 k
k=1 k + 5k
converges, too. As with the Integral Test, although the Comparison Test tells us that both
series converge, the two series need not converge to the same sum. A quick calculation
∞ 1
of a few partial sums should convince you that
converges to approximately
3
∞
k=1 k
1
converges to approximately 0.2798. (Note that this is consis1.202, while
3
k=1 k + 5k
tent with what we saw in Figure 8.26.)
for all k ≥ 1. Since
■
2.0 108
1.5 108
Example 3.6
1.0 108
Using the Comparison Test for a Divergent Series
Investigate the convergence or divergence of
0.5 108
n
5
10
15
Figure 8.27
n 5k + 1
Sn =
.
k
k=1 2 − 1
20
∞ 5k + 1
.
k
k=1 2 − 1
Solution
From the graph of the first 20 partial sums seen in Figure 8.27, it appears
that the partial sums are growing very rapidly. On this basis, we would conjecture that
the series diverges. Of course, to verifythis,
that for k
we need further testing.
Notice
∞
5 k
5 k
5k
large, the general term looks like k =
and we know that
is a divergent
2
2
k=1 2
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Section 8.3
The Integral Test and Comparison Tests
5
geometric series |r| = > 1 . Further,
2
5k + 1
5k
5k
≥
≥
=
2k − 1
2k − 1
2k
By the Comparison Test,
653
k
5
.
2
∞ 5k + 1
diverges, too.
k
k=1 2 − 1
■
There are plenty of series whose general term looks like the general term of a familiar
series, but for which it is unclear how to get the inequality required for the Comparison Test
to go in the right direction.
Example 3.7
Sn
A Comparison That Does Not Work
Investigate the convergence or divergence of the series
∞
k=3
k3
1
.
− 5k
0.16
Solution
Note that this is nearly identical to example 3.5, except that there is a
“−” sign in the denominator instead of a “+” sign. The graph of the first 20 partial sums
seen in Figure 8.28 looks somewhat similar to the graph in Figure 8.26, except that the
series appears to be converging to about 0.12. In this case, however, we have the
inequality
0.12
0.08
0.04
k3
n
5
10
15
20
Figure 8.28
n
1
.
Sn =
3
k=3 k − 5k
1
1
≥ 3 , for all k ≥ 3.
− 5k
k
∞ 1
Unfortunately, this inequality goes the wrong way: we know that
is a convergent
3
∞
k=3 k
1
p-series, but since
is “larger” than this convergent series, the Comparison
3
k=3 k − 5k
Test says nothing.
■
Think about what happened in example 3.7 this way: while you might observe that
k2 ≥
and you know that
“larger” series
∞
1
, for all k ≥ 1,
k3
∞ 1
is convergent, the Comparison Test says nothing about the
3
k=1 k
k 2 . In fact, we know that this last series is divergent (by the kth-term
k=1
test for divergence, since lim k 2 = ∞ = 0). To resolve this difficulty for the present
k→∞
problem, we will need to either make a more appropriate comparison or use the Limit
Comparison Test, which follows.
NOTES
When we say lim
k→∞
ak
= L > 0, we
bk
mean that the limit exists and is
positive. In particular, we mean that
lim
k→∞
ak
= ∞.
bk
Theorem 3.4 (Limit Comparison Test)
ak
= L > 0. Then,
Suppose that ak , bk > 0 and that for some (finite) value, L , lim
k→∞ bk
∞
∞
ak and
bk both converge or they both diverge.
either
k=1
k=1
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Chapter 8 Infinite Series
Proof
ak
ak
= L > 0, this says that we can make
as close to L as desired. So, in
bk
bk
L
ak
particular, we can make
within distance of L . That is, for some number N > 0,
2
bk
If lim
k→∞
L−
L
ak
L
< L + , for all k > N
<
2
bk
2
or
3L
L
ak
<
<
.
2
bk
2
(3.6)
Multiplying inequality (3.6) through by bk (recall that bk > 0), we get
3L
L
bk < ak <
bk , for k ≥ N.
2
2
L
bk =
ak converges, then the “smaller” series
Note that this says that if
2
k=1
∞
∞ k=1
L bk must also converge, by the Comparison Test. Likewise, if
ak diverges, the
2 k=1
k=1
∞
∞
∞
3L
3L bk =
bk must also diverge. In the same way, if
bk
“larger” series
2 k=1
k=1 2
k=1
∞
∞
3L
bk converges and so, too must the “smaller” series
ak .
converges, then
2
k=1
k=1
∞
∞
∞
L
bk diverges and hence, the “larger” series
bk diverges, then
ak
Finally, if
2
k=1
k=1
k=1
must diverge, also.
∞
∞
■
We can now use the Limit Comparison Test to test the series from example 3.7 whose
convergence we have so far been unable to confirm.
Example 3.8
Using the Limit Comparison Test
Investigate the convergence or divergence of the series
∞
1
.
3 − 5k
k
k=3
Solution
Recall that we had already observed in example 3.7 that the general term
1
1
ak = 3
“looks like” bk = 3 , for k large. We then consider the limit
k − 5k
k
ak
1
1
1
1
1 = lim
lim
= lim ak
= 1 > 0.
= lim 3
5
k→∞ bk
k→∞
k→∞ (k − 5k)
k→∞
bk
1
−
k3
k2
∞ 1
is a convergent p-series ( p = 3 > 1), the Limit Comparison Test says that
3
k=1 k
∞
1
is also convergent, as we had originally suspected.
3
k=1 k − 5k
Since
■
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Section 8.3
The Integral Test and Comparison Tests
655
The Limit Comparison Test can be used to resolve convergence questions for a great
many series. The first step in using this (like the Comparison Test) is to find another series
(whose convergence or divergence is known) that “looks like” the series in question.
Example 3.9
Using the Limit Comparison Test
Investigate the convergence or divergence of the series
∞
k=1
k 2 − 2k + 7
.
k 5 + 5k 4 − 3k 3 + 2k − 1
Solution
The graph of the first 20 partial sums in Figure 8.29 suggests that the series converges to a limit of about 1.61. The following table of partial sums supports this
conjecture.
Sn
n
1.62
n
k=1
1.60
1.58
1.56
1.54
1.52
1.50
n
5
Sn =
Sn =
n
k=1
10
15
k2 − 2k + 7
k5 + 5k4 − 3k3 + 2k − 1
5
1.60522
10
1.61145
20
1.61365
50
1.61444
75
1.61453
100
1.61457
20
Figure 8.29
k 2 − 2k + 7
.
5
k + 5k 4 − 3k 3 + 2k − 1
1
k2
= 3 (since the terms with the
k5
k
largest exponents tend to dominate the expression, for large values of k). From the Limit
Comparison Test, we have
Notice that for k large, the general term looks like
ak
k 2 − 2k + 7
1
1
= lim 5
4
3
k→∞ bk
k→∞ k + 5k − 3k + 2k − 1
k3
lim
(k 2 − 2k + 7)
k3
k→∞ (k 5 + 5k 4 − 3k 3 + 2k − 1) 1
1
(k 5 − 2k 4 + 7k 3 )
5
k1 = lim 5
4
3
k→∞ (k + 5k − 3k + 2k − 1)
k5
= lim
= lim
k→∞
1−
1+
5
k
−
2
k
3
k2
+
+
7
k2
2
k4
−
1
k5
= 1 > 0.
∞ 1
is a convergent p-series ( p = 3 > 1), the Limit Comparison Test says that
3
k=1 k
∞
k 2 − 2k + 7
converges, also. Finally, now that we have established
5
4
3
k=1 k + 5k − 3k + 2k − 1
that the series is in fact, convergent, we can use our table of computed partial sums to
approximate the sum of the series as 1.61457.
Since
■
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Chapter 8 Infinite Series
EXERCISES 8.3
1.
Notice that the Comparison Test doesn’t always give
us information about convergence or divergence. If
∞
bk diverges, explain why you can’t
ak ≤ bk for each k and
k=1
∞
ak diverges.
tell whether or not
21.
3.
Explain why the Limit Comparison Test works. In
ak
= 1, explain how ak and bk comparticular, if lim
k→∞ bk
∞
∞
ak and
bk either both converge or
pare and conclude that
k=1
k=1
both diverge.
∞
ak
ak
= 0 and
k→∞ bk
k=1
converges, explain why you can’t tell whether or not
∞
bk converges.
In the Limit Comparison Test, if lim
k=1
4.
23.
In exercises 5–40, determine convergence or divergence of the
series.
∞
4
√
5.
3
k
k=1
k −11/10
6.
∞
k=0
11.
∞
k=1
13.
∞
k=2
15.
∞
k=1
17.
k+1
2
k + 2k + 3
4
2 + 4k
2
k ln k
2k
k3 + 1
∞
e1/ k
k=1
k2
8.
10.
25.
27.
24.
∞
k2 + 1
√
k5 + 1
k=0
26.
29.
∞
k=1
31.
33.
∞
∞
e− k
√
k
k=1
k
12.
∞
∞
k=1
14.
∞
k=2
16.
∞
k=0
18.
k3
4
(2 + 4k)2
k2
2
∞
ke−k
4 + e−k
k=1
1 + k2
4
√
3
k +1
∞
k+2
√
3
k5 + 4
k=0
∞
sin−1 (1/k)
k2
k=1
1
cos2 k
30.
sin k + 2
k2
32.
∞
k=1
34.
k
1
sin2 k
∞
e1/ k + 1
k3
k=1
∞
2 + cos k
k=1
∞
k+1
k+2
∞
k+1
3 +2
k
k=1
k
36.
∞
k 3 + 2k + 3
4 + 2k 2 + 4
k
k=0
38.
∞
k+1
2 +2
k
k=1
40.
∞
k+1
4 +2
k
k=1
In our statement of the Comparison Test, we required that
ak ≤ bk for all k. Explain why the conclusion would remain
true if ak ≤ bk for k ≥ 100.
42.
If ak > 0 and
∞
ak converges, prove that
k=1
∞
ak2 converges.
k=1
43.
Prove the following extension of the Limit Comparison Test: if
∞
∞
ak
bk converges, then
ak converges.
= 0 and
lim
k→∞ bk
k=1
k=1
44.
Prove the following extension of the Limit Comparison Test: if
∞
∞
ak
bk diverges, then
ak diverges.
= ∞ and
lim
k→∞ bk
k=1
k=1
3
k(ln k)2
√
k
k2 + 1
28.
2k 2
+2
k 5/2
41.
k2 + 1
+ 3k + 2
∞ √
1 + 1/k
k=1
20.
39.
∞
k=0
∞
k 4 + 2k − 1
5 + 3k 2 + 1
k
k=1
k=1
∞
k=1
∞
ln k
k=2
37.
k 3/2
∞
tan−1 k
k=1
−9/10
∞
4
√
k
k=1
k=0
√
19.
∞
k=1
k=1
9.
2
√
2
k +4
∞
k=1
A p-series converges if p > 1 and diverges if p < 1.
What happens for p = 1? If your friend knows that the
harmonic series diverges, explain an easy way to remember the
rest of the conclusion of the p-series test.
∞
22.
k=0
35.
7.
3k
+2
k=1
k=1
2.
∞
In exercises 45–48, determine all values of p for which the series
converges.
45.
∞
k=2
1
k(ln k) p
46.
∞
k=0
1
, a > 0, b > 0
(a + bk) p
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Section 8.3
47.
∞
ln k
48.
kp
k=2
∞
k p−1 ek
p
49.
S100,
51.
S50,
53.
S40,
∞
1
4
k
k=1
∞
6
8
k
k=1
∞
ke−k
50.
S100,
52.
S80,
Prove that the every-other-term harmonic series 1 + 13 + 15 +
∞
1
1
+ · · · diverges. (Hint: Write the series as
and use
7
k=0 2k + 1
the Limit Comparison Test.)
62.
Would the every-third term harmonic series 1 + 14 + 17 +
1
+ · · · diverge? How about the every-fourth-term harmonic
10
series 1 + 15 + 19 + 131 + · · ·? Make as general a statement as
possible about such series.
∞
4
2
k
k=1
∞
k=1
2
54.
S200,
k=1
2
k2 + 1
63.
∞
tan−1 k
k=1
1 + k2
In exercises 55–58, determine the number of terms needed to
obtain an approximation accurate to within 10−6 .
55.
57.
∞
3
4
k
k=1
∞
ke−k
2
56.
∞
2
2
k
k=1
58.
∞
4
5
k
k=1
k=1
64.
In exercises 59 and 60, answer with “converges” or “diverges” or
“can’t tell.” Assume that ak > 0 and bk > 0.
59.
Assume that
∞
ak converges and fill in the blanks.
k=1
(a)
If bk ≥ ak for k ≥ 10, then
∞
bk
.
k=1
(b)
(c)
(d)
60.
∞
bk
= 0, then
bk
k→∞ ak
k=1
∞
bk
If bk ≤ ak for k ≥ 6, then
If lim
If lim
k→∞
Assume that
∞
∞
(b)
(c)
If bk ≥ ak for k ≥ 10, then
∞
∞
bk
.
k=1
bk
= 0, then
bk
ak
k=1
∞
bk
If bk ≤ ak for k ≥ 6, then
If lim
k→∞
.
.
k=1
(d)
If lim
k→∞
66.
∞
1
6
k
k=1
67.
∞
1
8
k
k=1
68.
∞
1
10
k
k=1
.
k=1
(a)
∞
1
4
k
k=1
69.
ak diverges and fill in the blanks.
∞
bk
= ∞, then
bk
ak
k=1
.
Estimate ζ(2) numerically. Compare your result with that of
exercise 56 of section 8.2.
65.
k=1
bk
= ∞, then
bk
ak
k=1
∞ 1
for
x
k=1 k
x > 1. Explain why the restriction x > 1 is necessary. Leonhard Euler, considered to be one of the greatest and most prolific mathematicians ever, proved the remarkable result that
1
.
ζ(x) =
1
p prime 1 − p x
The Riemann-zeta function is defined by ζ(x) =
In exercises 65–68, use your CAS or graphing calculator to
numerically estimate the sum of the convergent p-series and
identify x such that the sum equals ζ (x) for the Riemann-zeta
function of exercise 63.
.
.
657
61.
k=1
In exercises 49–54, estimate the error in using the indicated partial sum Sn to approximate the sum of the series.
The Integral Test and Comparison Tests
70.
∞
1
Numerically investigate the p-series
and
0.9
k
k=1
∞
1
and for other values of p close to 1. Can you
1.1
k=1 k
distinguish convergent from divergent series numerically?
∞ 1
diverges. This is the “smallest”
k=2 k
1
1
p-series that diverges, in the sense that < p for p < 1.
k
k
∞
1
1
1
< . Show that
diverges and
Show that
k ln k
k
k=2 k ln k
∞
1
1
1
<
diverges and
. Find a
k ln k ln(ln k)
k ln k
k=2 k ln k ln(ln k)
∞
1
ak diverges and ak <
series such that
. Is
k ln k ln(ln k)
k=2
there a smallest divergent series?
You know that
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Chapter 8 Infinite Series
8.4
ALTERNATING SERIES
So far, we have focused our attention on positive term series, that is, series all of whose
terms are positive. Before we consider the general case, we spend some time in this section
examining alternating series, that is, series whose terms alternate back and forth from
positive to negative. There are several reasons for doing this. First, alternating series appear
frequently in applications. Second, alternating series are surprisingly simple to deal with
and studying them will yield significant insight into how series work.
An alternating series is any series of the form
∞
(−1)k+1 ak = a1 − a2 + a3 − a4 + a5 − a6 + · · · ,
k=1
where ak > 0, for all k.
Example 4.1
The Alternating Harmonic Series
Investigate the convergence or divergence of the alternating harmonic series
∞
(−1)k+1
k=1
k
=1−
1 1 1 1 1
+ − + − + ···.
2 3 4 5 6
Solution
The graph of the first 20 partial sums seen in Figure 8.30 suggests that
the series might converge to about 0.7. We now calculate the first few partial sums by
hand. Note that
Sn
1.0
0.8
S1 = 1,
0.6
S2 = 1 −
0.4
S3 =
0.2
n
5
10
15
20
S4 =
Figure 8.30
n (−1)k+1
.
Sn =
k
k=1
S5 =
S6 =
S7 =
S8 =
1
1
= ,
2
2
1 1
5
+ = ,
2 3
6
5 1
7
− =
,
6 4
12
7
1
47
+ =
,
12 5
60
47 1
37
− =
,
60 6
60
37 1
319
+ =
,
60 7
420
319 1
533
− =
420 8
840
and so on. We have plotted these first 8 partial sums on the number line shown in
Figure 8.31.
S2
0.5
S4 S6 S8
0.6
S7 S5
0.7
0.8
S3
S1
0.9
Figure 8.31
∞ (−1)k+1
.
Partial sums of
k
k=1
1
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Section 8.4
n
Sn =
n (−1) k+1
k
k=1
Alternating Series
659
You should notice that the partial sums are bouncing back and forth, but seem to be
zeroing-in on some value. (Could it be the sum of the series?) This should not be surprising, since as each new term is added or subtracted, we are adding or subtracting
less than we subtracted or added (why did we reverse the order here?) to get the previous partial sum. You should notice this same zeroing-in in the accompanying table
displaying the first 20 partial sums of the series. Based on the behavior of the partial
sums, it is reasonable to conjecture that the series converges to some value between
0.66877 and 0.71877. We can resolve the question of convergence definitively with the
theorem that follows.
1
1
2
0.5
3
0.83333
4
0.58333
5
0.78333
6
0.61667
7
0.75952
8
0.63452
9
0.74563
10
0.64563
11
0.73654
12
0.65321
13
0.73013
14
0.65871
15
0.72537
16
0.66287
17
0.7217
18
0.66614
19
0.71877
a3
20
0.66877
a4
■
Theorem 4.1 (Alternating Series Test)
Suppose that lim ak = 0 and 0 < ak+1 ≤ ak for all k ≥ 1. Then, the alternating series
k→∞
∞
(−1)k+1 ak converges.
k=1
Before considering the proof of the theorem, make sure that you have a clear idea of
what it is saying. In the case of an alternating series satisfying the hypotheses of the theorem, we start with 0 and add a1 > 0 to get the first partial sum S1 . To get the next partial
sum, S2 , we subtract a2 from S1 , where a2 < a1 . This says that S2 will be between 0 and S1 .
We illustrate this situation in Figure 8.32.
a1
a2
a5
a6
0
S2
S4
S6
S5
S3
S1
Figure 8.32
Convergence of the partial
sums of an alternating series.
Continuing in this fashion, we add a3 to S2 to get S3 . Since a3 < a2 , we must have that
S2 < S3 < S1 . Referring to Figure 8.32, notice that
S2 < S4 < S6 < · · · < S5 < S3 < S1 .
In particular, this says that all of the odd-indexed partial sums (i.e., S2n+1 , for n = 0, 1, 2, . . .)
are larger than all of the even-indexed partial sums (i.e., S2n , for n = 1, 2, . . .). As the partial
sums oscillate back and forth, they should be drawing closer and closer to some limit S,
somewhere between all of the even-indexed partial sums and the odd-indexed partial sums,
S2 < S4 < S6 < · · · < S < · · · < S5 < S3 < S1 .
(4.1)
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Chapter 8 Infinite Series
Proof
Notice from Figure 8.32 that the even- and odd-indexed partial sums seem to behave
somewhat differently. First, we consider the even-indexed partial sums. We have
S2 = a1 − a2 > 0
and
S4 = S2 + (a3 − a4 ) ≥ S2 ,
since (a3 − a4 ) ≥ 0. Likewise, for any n, we can write
S2n = S2n−2 + (a2n−1 − a2n ) ≥ S2n−2 ,
since (a2n−1 − a2n ) ≥ 0. This says that the sequence of even-indexed partial sums
{S2n }∞
n=1 is increasing (as we saw in Figure 8.32). Further, observe that
0 < S2n = a1 + (−a2 + a3 ) + (−a4 + a5 ) + · · · + (−a2n−2 + a2n−1 ) − a2n ≤ a1 ,
for all n, since every term in parentheses is negative. Thus, {S2n }∞
n=1 is both bounded
(by a1 ) and monotonic (increasing). By Theorem 1.4, {S2n }∞
must
be convergent to
n=1
some number, say L .
Turning to the sequence of odd-indexed partial sums, notice that we have
S2n+1 = S2n + a2n+1 .
From this, we have
lim S2n+1 = lim (S2n + a2n+1 ) = lim S2n + lim a2n+1 = L + 0 = L ,
n→∞
n→∞
n→∞
n→∞
since lim an = 0. Since both the sequence of odd-indexed partial sums {S2n+1 }∞
n=0 and
n→∞
∞
the sequence of even-indexed partial sums {S2n }n=1 converge to the same limit, L , we
have that
lim Sn = L ,
n→∞
also.
■
Example 4.2
Using the Alternating Series Test
Reconsider the convergence of the alternating harmonic series
Solution
∞ (−1)k+1
.
k
k=1
Notice that
lim ak = lim
k→∞
k→∞
1
= 0.
k
Further,
0 < ak+1 =
1
1
≤ = ak , for all k ≥ 1.
k+1
k
By the Alternating Series Test, the series converges. (You can use the calculations from
example 4.1 to arrive at an approximate sum.)
■
The Alternating Series Test is certainly the easiest test we’ve discussed so far for determining the convergence of a series. It’s straightforward, but you will sometimes need to
work a bit to verify the hypotheses.
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Section 8.4
Example 4.3
Alternating Series
661
Using the Alternating Series Test
Investigate the convergence or divergence of the alternating series
∞ (−1)k (k + 3)
.
k(k + 1)
k=1
Solution
The graph of the first 20 partial sums seen in Figure 8.33 suggests that
the series converges to some value around −1.5. The following table showing some
select partial sums suggests the same conclusion.
Sn
n
5
10
15
20
Sn
n
0.5
1.0
1.5
2.0
Figure 8.33
n (−1)k (k + 3)
.
Sn =
k(k + 1)
k=1
n (−1) k (k + 3)
k(k + 1)
k=1
Sn =
n
n (−1) k (k + 3)
k(k + 1)
k=1
50
−1.45545
51
−1.47581
100
−1.46066
101
−1.47076
200
−1.46322
201
−1.46824
300
−1.46406
301
−1.46741
400
−1.46448
401
−1.46699
We can verify that the series converges by first checking that
(k + 3)
k→∞ k(k + 1)
lim ak = lim
k→∞
1
k2
1
k2
= lim
k→∞
1
k
+
1+
3
k2
1
k
= 0.
Next, consider the ratio of two consecutive terms:
(k + 4)
ak+1
k(k + 1)
k 2 + 4k
=
= 2
< 1,
ak
(k + 1)(k + 2) (k + 3)
k + 5k + 6
for all k ≥ 1. From this, it follows that ak+1 < ak , for all k ≥ 1 and so, by the Alternating Series Test, the series converges. Finally, from the preceding table, we can see that
the series converges to a sum between −1.46448 and −1.46699. (How can you be sure
that the sum is in this interval?)
■
Example 4.4
A Divergent Alternating Series
Determine whether the alternating series
Solution
∞ (−1)k k
converges or diverges.
k=1 k + 2
First, notice that
lim ak = lim
k→∞
k→∞
k
= 1 = 0.
k+2
So, this alternating series is divergent, since by the kth term test for divergence, the
terms must tend to zero in order for the series to be convergent.
■
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Chapter 8 Infinite Series
Estimating the Sum of an Alternating Series
We have repeatedly remarked that once you know that a series converges, you can always
approximate the sum of the series by computing some partial sums. However, in finding an
approximate sum of a convergent series, how close is close enough? Realize that answers
to such questions of accuracy are not “one size fits all,” but rather are highly contextsensitive. For instance, if the sum of the series is to be used to find the angle from the
ground at which you throw a ball to a friend, you might accept one answer. On the other
hand, if the sum of that same series is to be used to find the angle at which to aim your
spacecraft to ensure a safe reentry into the earth’s atmosphere, you would likely insist on
greater precision (at least, we would).
So far, we have calculated approximate sums of series by observing that a number of
successive partial sums of the series are within a given distance of one another. The underlying assumption here is that when this happens, the partial sums are also within that same
distance of the sum of the series. Unfortunately, this is simply not true, in general (although
it is true for some series). What’s a mathematician to do? For the case of alternating series,
we are quite fortunate to have available a simple way to get a handle on the accuracy. Note
that the error in approximating the sum S by the nth partial sum Sn is S − Sn .
Take a look back at Figure 8.32. Recall that we had observed from the figure that all of
∞
(−1)k+1 ak lie
the even-indexed partial sums Sn of the convergent alternating series,
k=1
below the sum S, while all of the odd-indexed partial sums lie above S. That is, [as in
(4.1)],
S2 < S4 < S6 < · · · < S < · · · < S5 < S3 < S1 .
This says that for n even,
Sn ≤ S ≤ Sn+1 .
Subtracting Sn from all terms, we get
0 ≤ S − Sn ≤ Sn+1 − Sn = an+1 .
Since an+1 > 0, we have
−an+1 ≤ 0 ≤ S − Sn ≤ an+1 ,
or
|S − Sn | ≤ an+1 , for n even.
(4.2)
Similarly, for n odd, we have that
Sn+1 ≤ S ≤ Sn .
Again subtracting Sn , we get
−an+1 = Sn+1 − Sn ≤ S − Sn ≤ 0 ≤ an+1
or
|S − Sn | ≤ an+1 , for n odd.
(4.3)
Since (4.2) and (4.3) (these are called error bounds) are the same, we have the same error
bound whether n is even or odd. This establishes the following result.
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Section 8.4
Alternating Series
663
Theorem 4.2
Suppose that lim ak = 0 and 0 < ak+1 ≤ ak for all k ≥ 1. Then, the alternating series
k→∞
∞
(−1)k+1 ak converges to some number S and the error in approximating S by the nth
k=1
partial sum Sn satisfies
|S − Sn | ≤ an+1 .
(4.4)
Notice that this says that the absolute value of the error in approximating S by Sn does
not exceed an+1 (the absolute value of the first neglected term).
Example 4.5
Estimating the Sum of an Alternating Series
∞ (−1)k+1
Approximate the sum of the alternating series
by the 40th partial sum and
k4
k=1
estimate the error in this approximation.
Solution
We leave it as an exercise to show that this series is convergent. We then
approximate the sum by
S ≈ S40 = 0.9470326439.
From our error estimate (4.4), we have
|S − S40 | ≤ a41 =
1
= 3.54 × 10−7 .
414
This says that our approximation S ≈ 0.9470326439 is off by no more than
±3.54 × 10−7 .
■
A much more interesting question than the one asked in example 4.5 is the following.
For a given convergent alternating series, how many terms must we take in order to guarantee that our approximation is accurate to a given level? We use the same estimate of error
from (4.4) to answer this question, as in the following example.
Example 4.6
Finding the Number of Terms Needed
for a Given Accuracy
∞ (−1)k+1
, how many terms are needed to
k4
k=1
of the actual sum S?
For the convergent alternating series
guarantee that Sn is within 1 × 10−10
Solution
In this case, we want to find the number of terms n for which
|S − Sn | ≤ 1 × 10−10 .
From (4.4), we have that
|S − Sn | ≤ an+1 =
1
.
(n + 1)4
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Chapter 8 Infinite Series
So, we look for n such that
1
≤ 1 × 10−10 .
(n + 1)4
Solving for n, we get
1010 ≤ (n + 1)4 ,
so that
√
4
1010 ≤ n + 1
or
n≥
√
4
1010 − 1 ≈ 315.2.
So, if we take n ≥ 316, we will guarantee an error of no more than 1 × 10−10 . Using the
suggested number of terms, we get the approximate sum
S ≈ S316 = 0.947032829447,
which we now know to be correct to within 1 × 10−10 .
■
EXERCISES 8.4
1.
If∞ ak ≥ 0, explain in terms of partial sums ∞why
(−1)k+1 ak is more likely to converge than
ak .
k=1
2.
3.
9.
k=1
Explain why in Theorem 4.2 we need the assumption
that ak+1 ≤ ak . That is, what would go wrong with the
proof if ak+1 > ak?
The Alternating Series Test was stated for the series
∞
(−1)k+1 ak . Explain the difference between
k=1
∞
∞
(−1)k ak and
(−1)k+1 ak and explain why we could have
k=1
k=1
∞
(−1)k ak .
stated the theorem for
11.
A common mistake is to think that if lim ak = 0, then
k→∞
∞
ak converges. Explain why this is not true for
In exercises 5–30, determine if the series is convergent or
divergent.
∞
3
(−1)k+1
5.
k
k=1
7.
∞
2
(−1)k 2
k
k=1
∞
(−1)k+1
6.
k=1
8.
4
k+1
∞
4
(−1)k √
k
k=1
k2
k
+2
10.
∞
2k 2 − 1
(−1)k+1
k
k=1
12.
∞
2k − 1
(−1)k
k3
k=1
13.
∞
k
(−1)k+1 k
2
k=1
14.
∞
3k
(−1)k+1
k
k=1
15.
∞
4k
(−1)k 2
k
k=1
16.
∞
k+2
(−1)k k
4
k=1
17.
∞
2k
k
+1
k=1
18.
19.
∞
3
(−1)k √
k
+1
k=1
20.
∞
k+1
(−1)k 3
k
k=1
21.
∞
2
(−1)k+1
k!
k=1
22.
∞
k!
(−1)k+1 k
3
k=1
23.
∞
k!
(−1)k k
2
k=1
24.
∞
4k
(−1)k
k!
k=1
k=1
positive-term series. This is also not true for alternating series
unless you add one more hypothesis. State the extra hypothesis
and explain why it’s needed.
∞
(−1)k
k=1
k=1
4.
∞
k2
(−1)k+1
k+1
k=1
25.
∞
(−1)k+1 2e−k
k=0
∞
k=1
26.
4k 2
k 2 + 2k + 2
∞
(−1)k+1 3e1/ k
k=1
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Section 8.4
27.
∞
(−1)k ln k
28.
k=2
29.
∞
1
(−1)k+1 k
2
k=0
30.
∞
1
(−1)k
ln
k
k=2
∞
(−1)k+1 2k
49.
A person starts walking from home (at x = 0) toward a friend’s
house (at x = 1). Three-fourths of the way there, he changes
his mind and starts walking back home. Three-fourths of the
way home, he changes his mind again and starts walking back
to his friend’s house. If he continues this pattern of indecision,
always turning around at the three-fourths mark, what will be
the eventual outcome? A similar problem appeared in a national
magazine and created a minor controversy due to the ambiguous wording of the problem. It is clear that the first
turnaround
is at x = 34 and the second turnaround is at 34 − 34 34 = 163 . But
is the third turnaround three-fourths of the way to x = 1 or
x = 34 ? The magazine writer assumed the latter. Show that with
this assumption, the person’s location forms a geometric series.
Find the sum of the series to find where the person ends up.
50.
If the problem of exercise 49 is interpreted differently, a more
interesting answer results. As before, let x1 = 34 and x2 = 163 .
If the next turnaround
of the way from x2 to 1,
is three-fourths
then x3 = 163 + 34 1 − 163 = 34 + 14 x2 = 51
. Three-fourths of
64
the way back to x = 0 would put us at x4 = x3 − 34 x3 =
1
51
x = 264
. Show that if n is even, then xn+1 = 34 + 14 xn and
4 3
1
xn+2 = 4 xn+1 . Show that the person ends up walking back and
forth between two specific locations.
51.
Use your CAS or calculator to find the sum of the alternating
∞
1
(−1)k+1 accurate to six digits. Compare
harmonic series
k
k=1
your approximation to ln 2.
52.
In this exercise, you
1 will determine whether or not the
improper integral 0 sin(1/x) dx converges. Argue that
1/π
1/(2π )
1
sin(1/x) dx , 1/(2π ) sin(1/x) dx , 1/(3π ) sin(1/x) dx , . . .
1/π
exist and that (if it exists),
1
1/π
1
sin(1/x) dx =
sin(1/x) dx +
sin(1/x) dx
In exercises 31–38, estimate the sum of each convergent series to
within 0.01.
∞
4
(−1)k+1 3
k
k=1
32.
∞
2
(−1)k+1 3
k
k=1
33.
∞
k
(−1)k k
2
k=1
34.
∞
k2
(−1)k k
10
k=1
35.
∞
3
(−1)k
k!
k=0
36.
∞
2
(−1)k+1
k!
k=0
37.
∞
4
(−1)k+1 4
k
k=1
38.
∞
3
(−1)k+1 5
k
k=1
In exercises 39–44, determine how many terms are needed to
estimate the sum of the series to within 0.0001.
39.
∞
2
(−1)k+1
k
k=1
40.
∞
2k
(−1)k
41.
k!
k=0
43.
45.
46.
47.
∞
4
(−1)k+1 √
k
k=1
∞
10 k
(−1)k
42.
k!
k=0
∞
k!
(−1)k+1 k
k
k=1
44.
∞
4k
(−1)k+1 k
k
k=1
In the text, we showed you one way to verify that a sequence is
decreasing. As an alternative, explain why if ak = f (k) and
f (k) < 0, then the sequence ak is decreasing. Use this method
k
is decreasing.
to prove that ak = 2
k +2
k=1
series can diverge even if lim ak = 0.
k→∞
48.
0
∞
1
= 1 − 13 + 15 − 17 + · · ·
2k + 1
converges. It can be shown that the sum of this series is π4 .
Verify that the series
k=0
(−1)k
+
k
Use the method of exercise 45 to prove that ak = k is
2
decreasing.
In this exercise, you will discover why the
Alternating Series
1/k if k is odd
,
Test requires that ak+1 ≤ ak . If ak =
1/k 2 if k is even
∞
k+1
(−1) ak diverges to ∞. Thus, an alternating
argue that
665
Given this result, we could use this series to obtain an approximation of π. How many terms would be necessary to get eight
digits of π correct?
k=0
31.
Alternating Series
1/(2π )
1/π
1/(2π )
1/(3π )
sin(1/x) dx + · · ·
Verify that the series is an alternating series and show that the
hypotheses of the Alternating Series Test are met. Thus, the
series and the improper integral both converge.
53.
∞
xk
, where x is a constant.
k
k=1
Show that the series converges for x = 1/2; x = −1/2; any x
such that −1 < x ≤ 1. Show that the series diverges if x = −1,
x < −1 or x > 1. We see in section 8.6 that when the series
converges, it converges to ln(1 + x). Verify this numerically
for x = 1/2 and x = −1/2.
Consider the series
(−1)k+1